MULTIPLE INTEGRATION NOTES AND EXERCISES. Antony Foster

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1 MULTIPL INTGRATION NOTS AN XRCISS by Antony Foster Overview In this discussion we consider the integral of a function of two variables z f(x, y) over a region in the xy-plane (-space) and the integral of a function of three variables w F(x, y, z) over a solid region in 3-space. These integrals are called multiple integrals and are defined as the limit of approximating Riemann sums, much like the single-variable integrals you learned about in earlier calculus courses. We can use multiple integrals to calculate quantities that vary over two or three dimensions, such as the total mass or the angular momentum of an object of varying density and the volumes of solids with general curved boundaries. OUBL INTGRATION In your earlier studies of calculus we defined the definite integral of a continuous function y f(x) over a closed interval [a, b] as a limit of Riemann sums. In this discussion we extend this idea to define the integral of a continuous function of two variables z f(x, y) over a bounded region in the plane. In both cases the integrals are limits of approximating Riemann sums. The Riemann sums for the integral of a single-variable function y (x) are obtained by: 1. Partitioning a finite interval [a, b] into thin subintervals, [x, x 1 ], [x 1, x ], [x, x 3 ],..., [x i 1, x i ],..., [x n 1, x n ] where a x and x n b multiplying the width of each subinterval, x i, by the value of f at a point x i. Forming a Riemann Sum by adding together all the products of the form f(x i ) x i. inside that subinterval, and then A similar method of partitioning, multiplying, and summing is used to construct double integrals. However, this time we pack a planar region with small rectangles, rather than small subintervals. We then take the product of each small rectangles area with the value of f(x, y) at a point inside that rectangle, and finally sum together all these products. When f(x, y) is continuous, these sums converge to a single number as each of the small rectangles shrinks in both width and height. The limit is the double integral of f(x, y) over. As with single integrals, we can evaluate multiple integrals via antiderivatives, which frees us from the formidable task of calculating a double integral directly from its definition as a limit of Riemann sums. The major practical problem that arises in evaluating multiple integrals lies in determining the limits of integration. While the integrals of your earlier studies of calculus were evaluated over an interval, which is determined by its two endpoints, multiple integrals are evaluated over a region in the plane or in space. This gives rise to limits of integration which often involve variables, not just constants. escribing the regions of integration is the main new issue that arises in the calculation of multiple integrals. OUBL INTGRALS OVR RCTANGULAR RGIONS OF TH xy-plan We begin our investigation of double integrals by considering the simplest type of planar region, a rectangle. We consider a function z f(x, y) defined on a rectangular region,, defined as {(x, y) : a x b, c y d}. IA: We subdivide the plane region into small rectangles using a network of lines parallel to the x- and y-axes. The lines divide into n rectangular pieces, where the number of such pieces n gets large as the width and height of each piece gets small. These rectangles form a partition of. A small rectangular piece of width, x and height y has area A x y. If we number the small pieces partitioning in some order, then their areas are given by numbers A 1, A,..., A n where A k is the area of the kth small rectangle. To form a Riemann sum over, we choose a point (x k, y k ) in the kth small rectangle, multiply the number f(x k, y k ) by the area A k and add together the products f(x k, y k ) A k : S n f(x k, y k ) A k k1 epending on how we pick (x k, y k ) in the kth small rectangle, we may get different values for S n.

2 We are interested in what happens to these Riemann sums as the widths and heights of all the small rectangles in the partition of approach zero. The norm of a partition P, written P is the largest width or height of any rectangle in the partition. For example, if P.1, then all the rectangles in the partition of have width at most.1 and height at most.1. Sometimes the Riemann sums converge as the norm of P goes to zero, written P. The resulting limit is then written as lim P k1 f(x k, y k ) A k. As P and the rectangles get narrow and short, their number n increases, so we can also write this limit as lim n k1 f(x k, y k ) A k. with the understanding that A k as n and P. There are many choices involved in a limit of this kind. The collection of small rectangles is determined by the grid of vertical and horizontal lines that determine a rectangular partition of P. In each of the resulting small rectangles there is a choice of an arbitrary point at which f is evaluated. These choices together determine a single Riemann sum. To form a limit, we repeat the whole process again and again, choosing partitions whose rectangle widths and heights both go to zero and whose number goes to infinity. When a limit of the sums exists, giving the same limiting value no matter what choices are made, then the function f(x, y) is said to be integrable over and the limit is called the double integral of f(x, y) over, written as f(x, y) da or f(x, y) dx dy. It can be shown that if f(x, y) is a continuous function throughout, then f is integrable, as in the single-variable case discussed in your earlier studies of calculus. Many discontinuous functions are also integrable, including functions which are discontinuous only on a finite number of points or smooth curves. The proof of these facts can be found in more advanced textbooks about calculus. Interpreting ouble Integrals as Volumes of Solids in 3-space When f(x, y) is a positive function over a plane region in the xy-plane, we may interpret the positive number f(x, y) da as the volume of the 3-dimensional solid region over the xy-plane bounded below by and bounded above by the graph of the surface given as z f(x, y). ach term f(x k, y k ) A k (a positive number) in the sum S n n k 1 f(x k, y k ) A k is the volume of a vertical rectangular box that approximates the volume of the portion of the solid region in space that stands directly above the base A k. The sum S n thus approximates what we want to call the total volume of the solid. efinition {Volume as a double Integral} If f(x, y) for all (x, y), then we define the Volume of the solid in space whose base is and bounded above by the surface z f(x, y) as V ( ), and V ( ) lim S n f(x, y) da n k1 where A k as n. As you might expect, this more general method of calculating volume agrees with the Solids of Revolution methods you encountered in your earlier studies of calculus, but we do not prove this here.

3 Fubini s Theorem for Calculating ouble Integrals Suppose that we wish to calculate the volume, V, under the plane z f(x, y) 4 x y over the rectangular region {(x, y) : x, y 1} in the xy-plane. If we apply the method of slicing you learned in working with volumes in your earlier studies of calculus, with slices perpendicular to the x-axis, then the volume is x x A(x) dx (1) where A(x) is the cross-sectional area at x. For each value of x, we may calculate A(x) as the integral A(x) y1 y (4 x y) dy () which is the area under the curve in the plane of the cross-section at x. In calculating A(x), x is held fixed and the integration takes place with respect to y. Combining quations (1) and (), we see that the volume of the entire solid is x x ( y1 ) Volume A(x) dx (4 x y) dy dx x x x x x [ 7 x x [4y xy y ( ) 7 x dx ] x x x y ] y1 dx y 5 cubic units To obtain the cross-sectional area A(x), we hold x fixed and integrate with respect to y. If we just wanted to write a formula for the volume, without carrying out any of the integrations, we could write Volume (4 x y) dy dx The expression on the right, called an iterated or repeated integral, says that the volume is obtained by integrating 4 x y with respect to y from y to y 1, holding x fixed, and then integrating the resulting expression in x with respect to x from x to x. The limits of integration and 1 are associated with y, so they are placed on the integral closest to dy. The other limits of integration, and, are associated with the variable x, so they are placed on the outside integral symbol that is paired with dx. What would have happened if we had calculated the volume by slicing with planes perpendicular to the y-axis? As a function of y, the typical cross-sectional area is A(y) x x (4 x y) dx ] x [4x x xy x 6 y. (4) The volume of the entire solid is therefore Volume y1 y A(y) dy y1 y (6 y) dy [ 6y y ] y1 y 5 cubic units, in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing Volume (4 x y) dx dy. The expression on the right says we can find the volume by integrating with respect to x from x to x as in quation (4) and then integrating the result with respect to y from y to y 1. In this iterated integral, the order of integration is first x and then y, the reverse of the order in quation (3).

4 What does the two volume calculations above with iterated integrals have to do with the double integral (4 x y) da over the rectangle {(x, y) : x, y 1}? The answer is that both iterated integrals give the value of the double integral. This is what we would reasonably expect, since the double integral measures the volume of the same region as the two iterated integrals. A theorem published in 197 by Guido Fubini says that: The double integral of any continuous function f(x, y) over a rectangle can be calculated as an iterated integral in either order of integration. Guido Fubini proved his theorem in greater generality, but this is what it says in our setting: Theorem 1. Fubini s Theorem (Weak Form) If f(x, y) is continuous throughout the rectangular region {(x, y) : a x b, c y d}, then f(x, y) da d b c a f(x, y) dx dy b d a c f(x, y) dy dx. Fubini s Theorem says that: ouble integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time. Fubini s Theorem also says that: We may calculate the double integral by integrating in either order. When we calculate a volume by slicing, we may use, either planes perpendicular to the x-axis or planes perpendicular to the y-axis. xample 1. Calculate f(x, y) da for f(x, y) 1 6x y and {(x, y) : x, 1 y 1}. Solution. By Fubini s Theorem f(x, y) da 1 1 Reversing the order of integration gives the same answer: 1 (1 6x y) dx dy [ x x 3 y ] x x dy ( 16y) dy [ y 8y ] y1 y (1 6x y) dy dx 1 [ y 3x y ] y1 y 1 dx [ (1 3x ) ( 1 3x ) ] dx dy 4

5 ouble Integrals over bounded Non Rectangular Regions To define the double integral of a function f(x, y) over a bounded, nonrectangular region, we again begin by covering with a grid of small rectangular cells whose union contains all points of. This time, however, we cannot exactly fill with a finite number of rectangles lying inside, since its boundary is curved, and some of the small rectangles in the grid lie partly outside. A partition of is formed by taking the rectangles that lie completely inside it, not using any that are either partly or completely outside. For commonly arising regions, more and more of is included as the norm of a partition (the largest width or height of any rectangle used) approaches zero. Once we have a partition of, we number the rectangles in some order from 1 to n and let A k be the area of the k-th rectangle. We then choose a point (x k, y k ) in the k-th rectangle and form the Riemann sum S n f(x k, y k ) A k k1 As the norm of the partition forming goes to zero, i.e., P, the width and height of each enclosed rectangle goes to zero and their number goes to infinity. If f(x, y) is a continuous function, then these Riemann sums converge to a limiting value, not dependent on any of the choices we made. This limit is called the double integral of f(x, y) over : lim f(x k, y k ) A k f(x, y) da P k1 The nature of the boundary of introduces issues not found in integrals over an interval. When has a curved boundary, the n rectangles of a partition lie inside but do not cover all of. In order for a partition to approximate well, the parts of covered by small rectangles lying partly outside must become negligible as the norm of the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made from polygons, circles, ellipses, and from continuous graphs over an interval, joined end to end. A curve with a fractal type of shape would be problematic, but such curves are not relevant for most applications. A careful discussion of which type of regions can be used for computing double integrals is left to a more advanced text. ouble integrals of continuous functions over nonrectangular regions have the same algebraic properties (summarized further on) as integrals over rectangular regions. The domain Additivity Property says that if is decomposed into nonoverlapping regions 1 and with boundaries that are again made of a finite number of line segments or smooth curves, then f(x, y) da f(x, y) da + f(x, y) da 1 If f(x, y) is positive and continuous over we define the volume of the solid region,, between and the surface z f(x, y) to be f(x, y) da as before. A Procedure for evaluating ouble Integrals over x-simple bounded nonrectangular Regions If {(x, y) : a x b, g 1 (x) y g (x)} is an x-simple region in the xy-plane that is, is bounded above and bounded below by the plane curves y g (x) and y g 1 (x) and on the sides by the lines x a and x b we may again calculate the volume f(x, y) da by the method of slicing. STP 1. We first calculate the cross-sectional area A(x) y g(x) y g 1(x) f(x, y) dy and then STP. integrate A(x) from x a to x b to get the volume as an iterated integral: Volume V ( ) f(x, y) da b A(x) dx b g(x) a a g 1(x) f(x, y) dy dx (5)

6 A Procedure for valuating ouble Integrals over y-simple bounded nonrectangular Regions Similarly, if {(x, y) : c x d, h 1 (y) x h (y)} is an y-simple region in the xy-plane that is, is bounded above and bounded below by the lines y c and y d and on the sides by the plane curves x h (y) and x h 1 (y), then the volume calculated by slicing as follows: Step 1. We first calculate the cross-sectional area A(y) x h(y) x h 1(y) f(x, y) dx and then Step. integrate A(y) from y c to y d to get the volume as an iterated integral: Volume V ( ) f(x, y) da d c A(y) dy d h(y) c h 1(y) f(x, y) dx dy (6) The fact that the iterated integrals (5) and (6) both give the volume that we defined to be the double integral of f(x, y) over is a consequence of the following stronger form of Fubini s Theorem. Theorem. Fubini s Theorem (Strong Form) Let f(x, y) be a continuous function throughout a region. 1. If is defined by a x b, g 1 (x) y g (x) with g 1 (x) and g (x) continuous on [a, b], that is, is an x-simple region, then f(x, y) da b g(x) a g 1(x) f(x, y) dy dx.. If is defined by c y d, h 1 (y) x h (y) with h 1 (y) and h (y) continuous on [c, d], that is, is a y-simple region, then f(x, y) da d h(y) c h 1(y) f(x, y) dx dy. xample. Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis (i.e., the line y ) and the lines y x and x 1 and whose top lies in the plane z f(x, y) 3 x y. Solution. First notice that the region is bounded below by the curve g 1 (x) and bounded above the the curve g (x) x and on the sides by the lines x and x 1, i.e, is an x-simple region. Thus by Fubini s Theorem (Strong Form): Volume f(x, y) da b g(x) a g 1(x) x f(x, y) da (3 x y) dy dx [3y xy y (3x 3 ) x dx ] yx y dx [ 3 x 1 ] x1 x3 x 1 cubic unit

7 When the order of the integration is reversed, i.e., we consider as a y-simple region, then the volume is Volume f(x, y) da d h(y) c h 1(y) y f(x, y) da (3 x y) dx dy [ 3x 1 x xy ] x1 xy dy (3 1 y 3y + 1 ) y + y dy ( 5 4y + 3 ) y dy [ 5 y y + 1 y3 ] y1 y 1 cubic unit. The two integrals are equal as they should be. Although Fubini s Theorem assures us that: A double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen. xample 3. Calculate f(x, y) da where f(x, y) sin(x) x by the x-axis, the line y x and the line x 1. Solution. [ x If we first integrate with respect to y and then with respect to x, we find sin(x) x ] dy dx [ y sin(x) ] yx x y dx If we reversed the order of integration and attempt to calculate and is the triangular region in the xy-plane bounded sin(x) dx [ cos(x)] x1 x 1 cos(1).46. y sin(x) x dx dy, we run into a problem because the integral sin(x) dx x cannot be expressed in terms of elementary functions (there is no simple antiderivative!) When calculating double Integrals over bounded nonrectangular regions, there is no general rule for predicting which order of integration will be the easiest or the good one in circumstances like these. If the order you choose first, doesnt work, then try the other. Sometimes neither order will work, and then we need to use numerical approximation methods.

8 A Procedure for Finding the limits of integration for a ouble Integrals We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works. When faced with evaluating f(x, y) da, integrating first with respect to y and then with respect to x, take the following steps: Step 1. Sketch: Sketch the region of integration and its bounding curves. Step. Find the y-limits of integration: Find the y-limits of integration imagining a vertical line, say L passing through the region in the direction of increasing y. Mark the y values where L enters and leaves. These are your y-limits of integration and are usually functions of x (instead of constants, unless is a rectangle). Step 3. Find the x-limits of integration: we find the x-limits of integration by choosing x-limits that include all vertical lines passing through. xercise 1. Write out a 3-step procedure similar to that above, when faced with evaluating f(x, y) da, integrating first with respect to x and then with respect to y. xample 4. {Reversing the order of integration} Reverse the order of integration for the double integral x f(x, y) da (4x + ) dy dx x using the above 3 Step procedure. Solution. f(x, y) da x y/ x (4x + ) dy dx 4 y (4x + ) dx dy

9 Properties of ouble Integrals Like single integrals, double integrals of continuous functions have algebraic properties that are useful in computations and applications. If f(x, y) and g(x, y) are continuous function on a plane region, then 1. Constant Multiple:. Sum and ifference: 3. omination: a) f(x, y) da if f(x, y) on. c f(x, y) da c f(x, y) da for any c. (f(x, y) ± g(x, y)) da f(x, y) da ± g(x, y) da. b) f(x, y) da g(x, y) da if f(x, y) g(x, y) on. 4. Additivity: f(x, y) f(x, y) da + g(x, y) da 1 if is the union of two nonoverlapping regions 1 and. Areas of bounded regions in the Plane If we take f(x, y) 1 in the definition of the double integral over a region in the preceding section, the Riemann sums reduce to S n f(x k, y k ) A k A k (1) k1 This is simply the sum of the areas of the small rectangles in the partition of, and approximates what we would like to call the area of the region. As the norm of a partition of approaches zero, the height and width of all rectangles in the partition approach zero, and the coverage of becomes increasingly complete. We define the area of the region, which we denote by A() be the limit Area of the region A() lim k1 P k1 A k da. efinition {area of planar region} The area of a closed, bounded region of the plane is A() da. Average Value The average value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length of the interval. For an integrable function of two variables defined on a bounded region in the plane, the average value is the integral of f over the region divided by the area of the region. This can be visualized by thinking of the function as giving the height at one instant of some water sloshing around in a tank whose vertical walls lie over the boundary of the region. The average height of the water in the tank can be found by letting the water settle down to a constant height. The height is then equal to the volume of water in the tank divided by the area of. We are led to define the average value of an integrable function f over a region to be efinition {Average value of f(x, y)} The average value of f(x, y) on a closed, bounded region of the plane is Avg(f) 1 f(x, y) da. A()

10 Moments, Mass and Centers-of-Mass for Thin Flat Plates (or Laminas) Using multiple integrals we can calculate the mass, Moment and center-of-mass (the balance point) of laminas with varying density. We first consider the problem of finding the center of mass of a thin flat plate or lamina: A disk of Aluminum, say, or a triangular sheet of metal. We assume the distribution of mass in such a plate to be continuous. A materials density function, denoted by δ(x, y) is its mass per unit area. The mass of a thin plate is obtained by integrating the density function δ(x, y) over the region forming the thin plate. The first moment about an axis is calculated by integrating over the distance from the axis times the density. The center of mass is found from the first moments. Below we give the double integral formulas for mass, first moments, and center of mass. Mass and first moment formulas for thin plates (also called laminas) covering a region in the xy-plane. MASS: M δ(x, y) da FIRST MOMNTS: M x y δ(x, y) da, CNTR-OF-MASS: x My M, ȳ Mx M M y x δ(x, y) da. xample 5. A thin plate covers a triangular region of the xy-plane bounded by the x-axis and the lines x 1 and y x in the first quadrant. The plate s density at the point (x, y) is δ(x, y) 6x + 6y + 6. Find the plate s Mass, First Moments and Center-of-Mass about the coordinate axes. Solution. 1. Make a sketch of {(x, y) x 1, y x} (Not shown). Calculating M δ(x, y) da by integrating first with respect to y and then with respect to x.. Finding the y-limits of integration: Any vertical line L passing through in the direction of increasing y must enter at y and leaves at y x and so A(x) x (6x + 6y + 6)dy [ 6xy + 3y + 6y ] yx y 4x + 1x 3. Finding the x-limits of integration: We include all vertical lines cutting through starting with x and ending with x 1. Thus [ x ] M δ(x, y) da (6x + 6y + 6) dy dx (4x + 1x) dx First Moment about the x-axis M x y δ(x, y) da [ x [ 8x 3 + 6x ] x1 x 14 mass Units ] (6xy + 6y + 6y) dy dx (8x 3 + 1x ) dx [ 7x 4 + 4x 3] x1 x 11 [ 3xy + y 3 + 3y ] yx y dx First Moment about the y-axis M y x δ(x, y) da [ x ] (6x + 6xy + 6x) dy dx (4x 3 + 1x ) dx [ 6x 4 + 4x 3] x1 x 1 [ 6x y + 3xy + 6xy ] yx y dx Finally, the coordinates of the Center-Of-Mass are: x M y M 1 14 and ȳ M x M In other words, the triangular plate balances at the point (5/7, 11/14) in.

11 ouble Integrals in Polar Form Integrals are sometimes easier to evaluate if we change to polar coordinates. This discussion shows how to accomplish the change and how to evaluate integrals over regions whose boundaries are given by polar equations. Integrals in Polar Coordinates When we defined the double integral of a function over a region in rectangular coordinates of the xy-plane, we began by cutting into rectangles whose sides were parallel to the x and y coordinate axes. These were the natural shapes to use because their sides have either constant x-values or constant y-values. In polar coordinates, the natural shape is a polar rectangle whose sides have constant r- and θ-values. Suppose that a function F(r, θ) is defined over a region that is bounded by the rays θ α and θ β and by the continuous curves r g 1 (θ) and r g (θ). Suppose also that g 1 (θ) g (θ) for every value of θ between α and β. Then lies in a fan-shaped region p defined by the inequalities r a and α θ β. We cover p by a grid of circular arcs and rays. The arcs are cut from circles centered at the origin, with radii, r, r,..., m r where r a m. The rays are given by: α, α + θ, α + θ,..., α + m θ where θ (β α) m. The arcs and rays partition p into small patches called polar rectangles. We number the polar rectangles that lie inside (the order does not matter), calling their areas: A 1, A,..., A n. We let (r k, θ k ) be any sample point in the polar rectangle whose area is A k. We then form the sum S n f(r k, θ k ) A k. k1 If F(r, θ) is continuous throughout, the above sum will approach a limit as we refine the grid to make r and θ go to zero. The limit is called the double integral of F(r, θ) over. In symbols, lim S n lim n n f(r k, θ k ) A k k1 F(r, θ) da. To evaluate the above limit, we first have to write the sum S n in a way that expresses A k in terms of r and θ. For convenience we choose r k to be the average of the radii of the inner and outer arcs bounding the k-th polar rectangle A k. The radius of the inner arc bounding is then r k ( r)/. The radius of the outer arc is r k + ( r)/. The area of a wedge-shaped sector of a circle having radius r and angle θ is Area of Sector 1 r θ (A formula you can derive using integration), as can be seen by multiplying π r, the area of the circle, by θ/π, the fraction of the circles area contained in the wedge. So the areas of the circular sectors subtended by these arcs at the origin are: Inner Radius: 1 Outer Radius: 1 ( r k r ) θ ( r k + r ) θ Therefore, A k area of large sector area of small sector [ ( θ r k + r ) ( r k r ) ] θ ( r k r) r k r θ

12 Combining this result with the sum defining S n gives S n F(r k, θ k ) A k k1 F(r k, θ k ) r k r θ k1 As n, and the values r and θ approach zero, these sums converge to the double integral lim S n lim n n F(r k, θ k ) r k r θ F(r, θ) r dr dθ k1 A version of Fubini s Theorem says that the limit approached by these sums can be evaluated by repeated single integrations with respect to r and θ as p F(r, θ) da θ β rg(θ) θ α rg 1(θ) p F(r, θ) r dr dθ. A Procedure For Finding Limits of Integration for Integrals in Polar coordinates The procedure for finding limits of integration in rectangular coordinates also works for polar coordinates. When faced with evaluating p F(r, θ) da over a region p in polar coordinates, integrating first with respect to r and then with respect to θ, take the following steps. Step 1. Sketch: Sketch the region of integration and its bounding curves. Step. Find the r-limits of integration: Find the r-limits of integration imagining a ray, say M passing through the region p in the direction of increasing r. Mark the r values where M enters and leaves p. These are your r-limits of integration. They usually depend on the angle θ that the ray M makes with the positive x-axis (that is, the r-limits are usually functions of θ (instead of constants, unless p is a circle)) Step 3. Find the θ-limits of integration: we find the θ-limits of integration by choosing θ-limits that include all rays passing through p. If F(r, θ) is the constant function whose value is 1 on p, then the integral of F over p is the area of p which we denote as A( p ). efinition {Area in Polar Coordinates} The area of a closed and bounded polar region p {(r, θ) α θ β, g 1 (θ) r g (θ)} in the polar coordinate plane is given by A( p ) r dr dθ. p or

13 Changing Cartesian (i.e., Integrals in Rectangular Coordinates) Integrals into Polar Integrals The procedure for changing a Cartesian integral f(x, y) da of the form f(x, y) dx dy, into a polar integral F(r, θ) r dr dθ requires two steps. p Step 1. First substitute r cosθ for x and r sinθ for y in the function f(x, y), and replace da dx dy by r dr dθ in the Cartesian integral. Step. Then supply polar limits of integration for the boundary of. The Rectangular Coordinates Integral then becomes f(x, y) dx dy f(r cosθ, r sinθ) r dr dθ p p F(r, θ) r dr dθ where F(r, θ) f(r cosθ, r sinθ) and p is the region of integration in Polar Coordinates. This is like the substitution method you encountered in your earlier studies of calculus, except that there are now two variables to substitute for instead of one. Notice that dx dy is not replaced by dr dθ but by r dr dθ instead. A more general discussion of changes of variables (substitutions) in multiple integrals is given in Section 16.9 of your textbook. xample 6. Find the polar moment of Inertia, I of a thin plate of density δ(x, y) 1 about the origin of a region bounded by the quarter circle x + y 1 in the first quadrant. Solution. The formula for I is I (x + y ) δ(x, y) da I x + I y. In rectangular coordinates the polar moment of Inertia is the value of the ouble integral (x + y )δ(x, y) da 1 x (x + y ) dy dx integrating first with respect to y and then with respect to x gives (x 1 x + (1 x ) 3/ ) 3 dx, an integral difficult to evaluate without tables. Things go better if we change the original integral to polar coordinates. Substituting r cosθ for x and r sin θ for y and rdrdθ for dy dx, we get 1 x (x + y ) dy dx π/ π/ [ r 4 4 (r ) r dr dθ ] r1 r dθ π/ 1 4 dθ π 8 Why is the polar coordinate transformation so effective here? One reason is that x + y simplifies to r. Another is that the limits of integration become constants.

14 Surface Area as a ouble Integral Idea: Let S be a surface whose equation is given by z f(x, y) with domain. We assume that f(x, y) is differentiable with continuous partial derivatives on a plane region of the xy-plane. We further assume that z f(x, y) on. Next, we partition into sub-rectangles R ij where 1 i m and 1 j n and whose area is A(R ij ) A x y. If we choose a sample point (x i, y j ) R ij at the corner of R ij closest to the origin (, ), then let P ij P ij (x i, y j, f(x i, y j )) be the point on the surface S directly above the sample point. The tangent plane to the surface S at the point P ij is an approximation to S near P ij. So the area T ij of the part of the tangent plane that lies directly above R ij is an approximation to the surface area S ij of the part of S that lies directly above R ij. Now the total area of S (the surface area) is approximated by the Riemann sum m T ij. i1 j1 Since the approximation to the total area of S improves as the number of sub-rectangles R ij increases, we therefore define the surface area of S to be m A(S) lim T ij. m,n i1 j1 For computational purposes, we need a more convenient formula for surface area than the one give above. In particular, we need a nice formula for T ij. The idea for a formula for T ij lies in our ability to parameterize the surface S whose equation is z f(x, y). Of course, we can parameterize z f(x, y) as follows: x x(x, y), y y(x, y), z f(x, y) for (x, y). Now we can write a vector function r(x, y) x(x, y)i + y(x, y)j + f(x, y)k with two parameters x and y as: r(x, y) x, y, f(x, y) for (x, y) {A parametric representation of z f(x, y)} We now have the vectors r x (x, y) and r y (x, y) that form the part of the tangent plane to the surface S at the point (x, y) and is given by Now r x (x, y) 1,, f x (x, y) and r y (x, y), 1, f y (x, y). i j k r x (x, y) r y (x, y) 1 f x (x, y) f x(x, y), f y (x, y), 1 1 f y (x, y) and the area of the tangent plane to the surface S at point (x, y) is given by r x (x, y) r y (x, y) ( f x (x, y)) + ( f y (x, y)) + 1 f x (x, y) + f y (x, y) + 1. Certainly, we see that the area of the surface S with equation z f(x, y) with (x, y), where f x (x, y) and f y (x, y) exists and are continuous is given by A(S) r x (x, y) r y (x, y) da f x (x, y) + f y (x, y) + 1 da {see Page 156 in your Textbook}. Clearly, based on our discussion above we see that a formula for T ij can be given as: T ij r x (x j, y j ) x r y(x i, y ) y r x (x i, y j ) r y(x i, y j ) A f x (x i, y j ) + f y (x i, y j ) + 1 A. This is what the author was saying on page 156 near the bottom of the page {My hope is that this derivation is clear}. m A(S) lim T ij m,n i1 j1 lim m m,n i1 j1 lim m m,n i1 j1 A(S) r x (x i, y j ) r y(x i, y j ) A f x (x i, y j ) + f y (x i, y j ) + 1 A f x (x, y) + f y (x, y) + 1 da.

15 Let us use our Surface Area formula above or see page 156 to solve some ouble Integration problems: {Study this worked out xercise carefully} xercise : Let us use double integrals to find a formula for the surface area of a sphere with radius R centered at the origin (,, ). Recall that the rectangular equation of such a sphere is given as x + y + z R where R is a positive constant. Solution. Solve for z to get z f(x, y) R x y {the upper hemisphere} and z g(x, y) f(x, y) R x y {the lower hemisphere}. Now to solve the problem we use only the upper hemisphere. Thus, x f x (x, y) and f y R y(x, y) so that x y Now f x (x, y) + f y (x, y) + 1 R x y x R x y + y R x y + R x y R x y A(S) R R x y da R R x y. where {(x, y) x +y R } is the projection of the upper hemisphere on the xy-plane. There is a subtle issue with this integral. The integrand in A(S) above {blows up! i.e., it goes to infinity} for points (x, y) on the boundary circle x + y R ; because of this, the double integral A(S) is an improper integral. Nevertheless, we can still compute A(S) provided that we approach the boundary circle as a limit from the interior towards the boundary. In addition, notice that f x (x, y) and f y (x, y) are not continuous on (the boundary of ) but are continuous on the interior of. Using Polar Coordinates we can evaluate the integral A(S) as follows: π α R A(S) lim r dr dθ α R R r ( π ) ( α R dθ lim α R ( ( πr) [ ] rα ) R r lim α R ) r R r dr r ( πr)( lim α R [ R α R ] ) 4πR {This formula should appear somewhere in your textbook}

16 Instructions For solving the problems: YOUR XAMINATION PAPR BGINS HR. It is important to Read these Notes First before solving any of the problems. 1. raw a labeled picture of the region of integration for each ouble Integral that you do.. Give a set description of the solid region of integration for each triple integral that you do. Also state whether the region of integration is x-simple, y-simple, r-simple or θ-simple. 3. Give a set description of the region of integration for each double integral that you do. Also state whether the region of integration is a Type I or II or III solid region. 4. ach Integral (ouble or Triple) require a 3 step or 4 step procedure described above. Show how you are using these steps. raw a sketch (of surfaces and regions of integration) where possible. 5. All problems are to be done in a neat and readable way! 6. This is an xam. The Rules Apply (on t forget that). o all the problems! PROBLM 1 {To be handed in}: Consider a right circular cylinder of radius R and height H (which includes the top, bottom and the sides) whose surface area A(S) is given as πr + πrh. where (S 1 {the top}, S {the bottom}, and S 3 {the side}) denotes various pieces of your cylinder S S 1 S S 3 Use double integrals A(S) A(S 1 ) + A(S 3 ) 1 f x (x, y) + f y (x, y) + 1 da to derive the above formula. You must give a description and a sketch of the plane regions 1 and ). is the projection of the graph of f(x, y) (which you must supply!) in the xy-plane. Hint: Sketch the right circular cylinder S so the either the x or y-axis acts as a central axis of rotation (or symmetry). Consider only the portion of the surface S 3 in the first octant. This should give you some idea of how to define the surface S 3 as a function f of two variables. Furthermore, you should carefully read through xercise above before taking on this problem!. PROBLM {To be handed in}: Use double integrals to derive the given formula for the surface area of a right circular cone of radius R and height H. The surface area of a cone is given by the formula πr + πr R + H. Hint: Consider the right circular cone surface with equation z f(x, y) H R x + y PROBLM 3 {To be handed in}: Use double integrals to derive the given formula for the surface area of a cap of a sphere of radius R and height H where < H < R. ( The cap of a sphere is the portion of the sphere bounded below by the plane z R H and bounded above by the plane z R). The surface area of a cap of a sphere with radius R is given by the formula πrh. Question: Consider what happens to the above formula as H approaches R (xplain). Is the result of this consideration what you expected? PROBLM 4 {To be handed in}: Use a double integral of the form f(x, y)da derive the given formula for the volume of the sphere x + y + z R described in xercise. The volume of such a sphere is given by the formula 4 3 πr3

17 PROBLM 5 {To be handed in}: Use double integrals to derive the given formula for the volume of a right circular cone of radius R and height H. The volume of a cone is given by the formula 1 3 πr H. PROBLM 6 {To be handed in}: Use a double integral of the form f(x, y) da (Pay close attention to how you write f(x, y)) to derive the given formula for the volume a of a cap of a sphere of radius R and height H where < H < R. ( The cap of a sphere is the portion of the sphere bounded below by the plane z R H and bounded above by the plane z R). The volume of a cap of a sphere with radius R is given by the formula 1 3 πh (3R H). Question: Consider what happens to the above formula as H approaches R (xplain?). Is the result of this consideration what you expected?

18 Worked out xercise 4 (on page 158) Find the area of the part of the paraboloid z f(x, y) 4 x y that lies above the plane z (i.e., the xy-plane) { See the picture below} Courtesy of MatLab Solution. We need to compute G(x, y) f x (x, y) + f y (x, y) + 1. So, f x (x, y) x and f y (x, y) y so that f x (x, y) + f y (x, y) + 1 4x + 4y + 1. Now is the projection of the surface S (i.e., the graph of z f(x, y)) on the xy-plane (z ), thus {(x, y) x + y 4} this a circular region centered at (, ) in R of radius r where r. We can now compute our surface area A(S) as follows: A(S) π G(x, y)da 4(x + y ) + 1dA π π f x (x, y) + f y (x, y) + 1 da 4r + 1 r dr dθ where {(r, θ) r ; θ π} in polar coordinates dθ 4r + 1 8r dr dθ 7 π 6 [( 17 ) 3 ( 1 ) 3] π 6 ( ) sq units. 1 [ ] u17 1 udu 8 π 3 ( u ) 3 u1

19 Worked Out xercise 1 { page 158} Find the surface area of the part of the sphere whose equation is x + y + z 4z that lies inside the paraboloid z x + y (Be Careful! This is not a cone!) Solution. Rewrite the equation of the sphere as (x ) + (y ) + (z ). We now have to find out how the paraboloid and the sphere intersect; i.e., we must solve the equation x + y + (z ) 4 z x y for z. So we obtain (z ) 4 z or z 3z and so z and z 3. This tells us that the two surfaces meet in the planes z (the bottom of the paraboloid) and z 3. The upper hemisphere (the top half of the sphere) whose equation is z f(x, y) + 4 x y is the part of the sphere that lies inside the paraboloid and they intersect in the plane z 3. The trace of the upper hemisphere in the plane z 3 is the boundary curve of the region in the z 3-plane. This boundary curve has equation x y or x + y 3 {This is a circle of radius 3}. Now {(x, y) x + y 3} is the projection of the Surface on the plane z 3. Next, x f x (x, y) and f 4 x y(x, y) y are continuous on and so our double integral below will y 4 x y not be an improper integral as it was in the case of xercise 1. Now G(x, y) f x (x, y) + f y (x, y) + 1 A(S) G(x, y)da 4 x y da π 3 π (1 ) 4π sq. units r 4 r 4 x y dr dθ

20 Triple Integrals in Rectangular Coordinates Just as double integrals allow us to deal with more general situations than could be handled by single integrals, triple integrals enable us to solve still more general problems. We use triple integrals to calculate the volumes of threedimensional shapes, the masses and moments of solids of varying density, and the average value of a function over a threedimensional region. Triple integrals also arise in the study of vector fields and fluid flow in three dimensions, as is discussed in Chapter 17. Triple Integrals IA: We extended the notion of a double integral over a region R to 3 dimensions (R 3 ). Let us consider the following simple set B R 3 defined as: B {(x, y, z) α x β, γ y δ, σ z τ} {called a solid rectangular box}. We next form a partition of the box B by dividing it into little sub-boxes by partitioning the interval [α, β] into l sub-intervals of equal width x, and dividing the interval [γ, δ] into m sub-intervals of equal width y and similarly dividing the interval [σ, τ] into n sub-interval of equal width z. The planes through the endpoints of these sub-intervals parallel to the coordinate planes divides B into lmn sub-boxes B ijk [x i 1, x i ] [y j 1, y j ] [z k 1, z k ] with 1 i l, 1 j m, 1 k n. ach sub-box B ijk has a volume V ijk x i y j z k. If F(x, y, z) is a function of three variables defined on B, then we form the Riemann triple sum m k1 j1 i1 l F(x ijk, y ijk, z ijk ) V ijk where (x ijk, y ijk, z ijk ) is a sample point in the sub-box B ijk in the ijk-th position of the partition of B. Now we define the Triple integral over the box B as the limiting value (if it exists) of the Riemann triple sums above; i.e., The triple integral of F over the box B is given by B F(x, y, z)dv lim l,m,n m k1 j1 i1 l F(x ijk, y ijk, z ijk ) V ijk. If F(x, y, z) is continuous on the box B then the sample point can be chosen to be any point (x k, y k, z k ) in the k-th sub-box B k (assuming we have n sub-boxes) and so the integral above can be simply written as B F(x, y, z)dv lim n F(x k, y k, z k ) V k (1). Fubini s Theorem xtended to R 3 : If w F(x, y, z) is continuous on the solid rectangular box k1 B {(x, y, z) α x β, γ y δ, σ z τ}, then B F(x, y, z)dv τ δ β σ γ α β τ δ α σ γ β δ τ α γ σ F(x, y, z)dxdy dz F(x, y, z)dy dz dx. F(x, y, z)dz dy dx There is a total of six possible orders of the above iterated triple integral (I wrote out 3 of them for you) PROBLM 7 {To be handed in}: Write out the other 3 orders of integration for the triple integral given above.

21 Integration over more general bounded regions (solids) of R 3 IA: We proceed in much the same way as we did above by enclosing the solid region inside the solid box B, thus B. We then define a new function G(x, y, z) which agrees with F(x, y, z) over the solid region but is for points over B that are not on ; that is, F(x, y, z) if (x, y, z) G(x, y, z) if (x, y, z) B \ The new function G(x, y, z) is called the extension of F(x, y, z) on to B. Thus, we have F(x, y, z)dv G(x, y, z)dv. Such an integral exists only if F(x, y, z) is a continuous function and (the boundary of ) is reasonably smooth. By this, we mean that we can parametrically represent the boundary surface of. The triple integral above now have the usual properties similar to those of double integrals. Naturally, of course, these solid regions of 3-space can have a variety of shapes, so we restrict our attention to continuous functions w F(x, y, z) and to certain types of simple regions. Integration over Type I solid Regions in Space We say a solid region is of Type I if it lies between the graphs of two continuous functions of x and y, that is, {(x, y, z) (x, y) xy, u 1 (x, y) z u (x, y)} where xy is the projection of in the xy-plane (z ). It follows immediately, that if is a Type I, then we can express the triple integral F(x, y, z) dv as F(x, y, z)dv xy B [ u(x,y) u 1(x,y) ] F(x, y, z) dz da. In particular, if xy is the projection of onto the xy-plane is an x-simple plane region, then {(x, y, z) α x β, g 1 (x) y g (x), u 1 (x, y) z u (x, y)} and the above integral becomes F(x, y, z)dv β g(x) [ u(x,y) α g 1(x) u 1(x,y) ] F(x, y, z) dz dy dx {if xy is an x-simple plane region} δ h(y) [ u(x,y) γ h 1(y) u 1(x,y) ] F(x, y, z) dz dxdy {if xy is a y-simple plane region}. Integration over Type II solid regions in space We say that a solid region is of Type II if it lies between two continuous functions of y and z, that is, {(x, y, z) (y, z) yz, u 1 (y, z) x u (y, z)} where the plane region yz is the projection of in the yz-plane (x ) and so we can express the triple integral F(x, y, z) dv as [ u(y,z) ] F(x, y, z)dv F(x, y, z) dx da. yz u 1(y,z)

22 PROBLM 8 {To be handed in}: Suppose is a Type II solid region in space. In particular, if the plane region yz, the projection of the solid region on the yz-plane is a y-simple plane region or an z-simple plane region, then write out two formulas {similar in form to the red formulas above} for F(x, y, z)dv yz [ u(y,z) u 1(y,z) ] F(x, y, z) dx da. Integration over Type III solid regions in Space PROBLM 9 {To be handed in}: efine a Type III solid region. {Hint: See the blue and red sections above}. PROBLM 1 {To be handed in}: Suppose is a Type III solid region in space and if your plane region, xz, the projection of the solid region on the xz-plane is an x-simple plane region or a z-simple plane region, then write out two formulas again for F(x, y, z)dv xz [ u(x,z) u 1(x,z) ] F(x, y, z) dy da.

23 Volume of a Bounded Solid Region in Space If F(x, y, z) is the constant function whose value is 1 on, then the Riemann sums in quation (1) above reduce to S n V k. k1 As x k, y k and z k approach, the cells V k becomes smaller and more numerous and fill up more and more of. We therefore define the volume of, denoted as V ( ) to be the triple integral V ( ) lim V k dv. n k1 If is a solid Type I region and w F(x, y, z) 1 everywhere on, then we write the volume of the solid as [ u(x,y) ] [ V ( ) dv dz da u (x, y) u 1 (x, y) ] da xy u 1(x,y) A Procedure For finding Limits of Integration For a TYP 1 Triple Integral We evaluate a triple integral by applying a three-dimensional version of Fubinis Theorem to evaluate it by three repeated single integrations. As with double integrals, there is a geometric procedure for finding the limits of integration for these single integrals. When faced with evaluating a Type 1 triple integral F(x, y, z) dv over a solid region in space, integrating first with respect to z, then with respect to y and finally with respect to x, take the following steps: Step 1. Sketch: Sketch the region of integration along with its shadow xy (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces of and the upper and lower bounding curves of xy. xy Step. Find the z-limits of integration: Find the z-limits of integration imagining a vertical line, say M passing through a typical point (x, y) in xy in the direction of increasing z. Mark the z values where the line M enters and leaves. These are your z-limits of integration. They usually depend on the point (x, y) where the line M enters ( at z u 1 (x, y))and leaves at z u (x, y)) (that is, the z-limits are usually functions of x and y (instead of constants, unless is a Box)) Step 3. Find the remaining limits of integration: we find the remaining limits of integration by using STPS and 3 of the Procedure for finding the limits of integration for the ouble Integral xy [u (x, y) u 1 (x, y)] da. This double integral is over an x-simple region xy, thus b g(x) F(x, y, z) dv [u (x, y) u 1 (x, y)] da [u (x, y) u 1 (x, y)] dy dx xy a g 1(x) Follow similar procedures if you change the order of integration. The shadow of region lies in the plane of the last two variables with respect to which the iterated integration takes place. xercise. Write out procedures similar to above when faced with valuating Triple Integrals over Type II and Type III regions in space (Note: You must first define Type III regions (see Problem 9)) Average Value of a Function in Space The average value of a function F(x, y, z) over a region solid region in space is defined by the formula Average Value of F(x, y, z) over 1 F(x, y, z) dv () V ( ) For example, if F(x, y, z) x + y + z, then the average value of F(x, y, z) over is the average distance of points in from the origin. If F(x, y, z) is the temperature at point (x, y, z) on a solid that occupies a region in space, then the average value of F(x, y, z) over is the average temperature of the solid.

24 Properties of Triple Integrals Triple integrals have the same algebraic properties as double and single integrals. If F(x, y, z) and G(x, y, z) are continuous function on a solid bounded region in space, then 1. Constant Multiple: c F(x, y, z) dv c F(x, y, z) dv for any c.. Sum and ifference: 3. omination: a) F(x, y, z) dv if F(x, y, z) on. (F(x, y, z) ± G(x, y, z)) dv F(x, y, z) dv ± G(x, y, z) dv. b) F(x, y, z) dv G(x, y, z) dv if F(x, y, z) G(x, y, z) on. 4. Additivity: F(x, y, z) F(x, y, z) dv + G(x, y, z) dv 1 if is the union of two nonoverlapping regions 1 and. Masses and Moments of solids in Three imensions If δ(x, y, z) is the density of an object occupying a bounded solid region in space (mass per unit volume), the integral of over gives the mass of the object. To see why, imagine partitioning the object into n mass elements m k δ(x k, y k, z k ) V k. The objects mass is the limit M lim n k1 m k lim n k1 δ(x k, y k, z k ) V k δ(x, y, z) dv We now derive a formula for the moment of inertia. If r(x, y, z) is the distance from the point (x, y, z) in to a line L, then the moment of inertia of the mass m k δ(x k, y k, z k ) V k about the line L is approximately I k r (x k, y k, z k ) m k. The moment of inertia about L of the entire object is I L lim n k1 I k lim n k1 r (x k, y k, z k ) δ(x k, y k, z k ) V k r δ(x, y, z) dv If L is the x-axis, then r y + z and I x (y + z ) δ dv. Similarly, if L is the y-axis or the z-axis, then we have I y (x + z ) δ dv and I z (x + y ) δ dv. Likewise, we can obtain the first moments about the coordinate planes. For example, gives the first moment about the yz-plane. x δ(x, y, z) dv The mass and moment formulas in space analogous to those discussed for planar regions in our earlier discussions above are summarized below.