MULTIPLE INTEGRALS. h 2 (y) are continuous functions on [c, d] and let f(x, y) be a function defined on R. Then



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MULTIPLE INTEGALS 1. ouble Integrals Let be a simple region defined by a x b and g 1 (x) y g 2 (x), where g 1 (x) and g 2 (x) are continuous functions on [a, b] and let f(x, y) be a function defined on. Then ) I = is called an iterated integral of f over. ˆ b (ˆ g2 (x) a g 1 (x) f(x, y)dy dx Similarly, if is a simple region defined by c y d and h 1 (y) x h 2 (y), where h 1 (y) and h 2 (y) are continuous functions on [c, d] and let f(x, y) be a function defined on. Then ) J = is called an iterated integral of f over. ˆ d (ˆ h2 (y) c h 1 (y) f(x, y)dx dy Notation. Sometimes I may be written as ) I = ˆ b (ˆ g2 (x) a g 1 (x) and J may be written as J = ˆ d (ˆ h2 (y) c h 1 (y) f(x, y)dy f(x, y)dx ) dx = dy = ˆ b ˆ g2 (x) a g 1 (x) ˆ d ˆ h2 (y) c h 1 (y) f(x, y) dy dx = f(x, y) dx dy = ˆ b a ˆ d c dx dy ˆ g2 (x) g 1 (x) ˆ h2 (y) h 1 (y) f(x, y) dy, f(x, y) dx. Example 1.1. Calculate both iterated integrals I and J of the function f(x, y) = 3xy 2 2x 2 y over where is bounded by x =, x = 2, y = 1 and y = 2. 1

2 MULTIPLE INTEGALS Theorem 1.1. Suppose that f(x, y) is continuous on, where is defined by a x b, g 1 (x) y g 2 (x) or c y d, h 1 (y) x h 2 (y). Then the double integral of f over, denoted by f(x, y)da, is f(x, y) da = ˆ b ˆ g2 (x) a g 1 (x) f(x, y) dy dx = ˆ d ˆ h2 (y) c h 1 (y) f(x, y) dx dy. Properties. Let f(x, y) and g(x, y) be two integrable functions over, a and b be two arbitrary constants, and = 1 2. Then (i) af(x, y) + bg(x, y) da = a f(x, y) da + b g(x, y) da. (i) f(x, y) da = f(x, y) da + f(x, y) da. 1 1 Example 1.2. (1) If = {(x, y) x 2 and 1 y 4}, evaluate (6x 2 + 4xy 3 ) da. (2) Let be the region bounded by y = x, y = and x = 4, write f(x, y) da as an iterated integral. (3) Evaluate x da, where is the portion of disk x 2 + y 2 16 in the second quadrant. (4) Write f(x, y) da as an iterated integral, where is the region bounded by x = y 2 and x = 2 y. (5) Let be the region bounded by the graphs of y = x, y = and x = 4. Evaluate (4e x2 5 sin y)da. (6) Let be the region bounded by the graphs of y = x, x = and y = 3. Evaluate (2xy 2 + 2y cos x) da. (7) Write and x 2 + 2y 2 = 1. f(x, y) da as an iterated integral, where is the region bounded by x 2 +y 2 = 1 Example 1.3. everse the order of the integration

(1) (2) (3) (4) (5) (6) ˆ 2 ˆ 2x 1 x ˆ 2 ˆ 4 y 2 ˆ 1 ˆ 2x x 2 ˆ x 2 ˆ 2x ˆ x 2 e ˆ ln y ˆ1 3 ˆ x f(x, y) dy dx. f(x, y) dx dy. 4 y 2 f(x, y) dy dx. f(x, y) dy dx. f(x, y) dx dy. f(x, y) dy dx + ˆ 6 ˆ 6 x Example 1.4. Evaluate the integral. (1) (2) (3) (4) (5) ˆ 1 ˆ 1 y ˆ 1 ˆ 1 ˆ 9 ˆ 3 1 e x2 dxdy. x e y3 dy dx. y e x2 2x x + 1 dx dy. 3 MULTIPLE INTEGALS 3 f(x, y) dy dx. x(x 1)e xy da where is bounded by the lines x =, y = and x + y = 2. (x + y) da where is bounded by the lines y = 2x, x = and y = 4. (6) (7) (8) (9) (1) ˆ 8 ˆ ln x ˆ2 π ˆ 1+cos y ˆ 1 ˆ 1 ˆ 1 ˆ 1 ˆ 2 ˆ x 1 1 x e y dy dx. y 2 x 2 sin y dx dy. dy dx. x 2 + 1 x (1 + x 2 + y 2 ) 3 2 y 2 dy dx. x2 dx dy. 2. Area The area of a bounded region in the real plane can be computed by da = Area of.

4 MULTIPLE INTEGALS For example, if has the shape then, A() = ˆ b ˆ g2 (x) a g 1 (x) dy dx. If has the shape then, A() = ˆ d ˆ h2 (y) c h 1 (y) dx dy. Example 2.1. Find the area of the region where is bounded by (1) y = x and y = x 2 in the first quadrant. (2) y = x 2 and y = x + 2. (3) xy = 1, y = x, x = e. (4) y = 1 4 x2 1 and y = 2 x. (5) x = y 2, y x = 3, y = 3, and y = 2. Example 2.2. Let 1 = {(x, y) x 2 + 2y 2 1}, 2 = {(x, y) 2x 2 + y 2 1} and = 1 2. Find the area of. The double integral 3. Volume f(x, y)da of a positive function f(x, y) can be interpreted as the volume under the surface z = f(x, y) over the region.

MULTIPLE INTEGALS 5 If f(x, y) g(x, y) for all (x, y) in, then the double integral (f(x, y) g(x, y))da is the volume between the surface z = f(x, y) and the surface z = g(x, y). Example 3.1. For example, the volume of the tetrahedron bounded by the plane 2x+y+z = 2 and the three coordinate planes (See the figure below) is given by V = ˆ 1 ˆ 2 2x (2 2x y) dy dx Example 3.2. Find the volume of the solid lying in the first octant and bounded by the graphs of z = 4 x 2, x + y = 2, x =, y = and z =. Example 3.3. (1) Find the volume of the solid bounded by the graphs of z = 2, z = x 2 +1, y = and x + y = 2. (2) A solid T is bounded by the coordinate planes, the plane z = x + 2y + 1 and the planes x = 1 and y = 2. Find the volume of T. (3) A solid T lies in the first octant and is bounded by the coordinate planes, and the planes x + 2y = 2 and x + 4y + 2z = 8. Find the volume of T.

6 MULTIPLE INTEGALS (4) A solid T is bounded by the elliptic paraboloid z = 1 4 x2 + 1 9 y2, the coordinate planes and the planes x = 2 and y = 3. Find the volume of T. (5) A solid T lies in the first octant and is bounded by the elliptic cylinder 4x 2 + z 2 = 1, the plane y = x, and the planes y = and z =. (6) Find the volume enclosed between the two surfaces z = x 2 + 3y 2 and z = 8 x 2 y 2. (7) A solid T is bounded above by z = 2 x 2 y 2 and below by z = x 2 + y 2. Find the volume of T. (8) A solid T is bounded by the sphere x 2 +y 2 +z 2 = 4 and the paraboloid x 2 +y 2 = 4(1 z). Find the volume of T. (9) Find the volume V of the solid region T where T is bounded by the sphere x 2 +y 2 +z 2 = 4 and the paraboloid x 2 + y 2 = 4(1 z). (1) A solid T is bounded by the paraboloids x 2 + y 2 z = and x 2 + y 2 + 2z = 1. Find the volume of T. (11) Find the volume of the solid in the first octant bounded by the circular paraboloid z = x 2 + y 2, the cylinder x 2 + y 2 = 4 and the coordinate planes. Find the volume of T. (12) A solid T is bounded by the cone z 2 = xy and the plane x + y = 4. Find the volume of T. (13) Use double integration to find the volume of the tetrahedron bounded by the coordinate planes and the plane 3x + 6y + 4z 12 =. (14) Find the volume of the solid lying in the first octant and bounded by z = 4 x 2, x + y = 2, x = and z =. (15) Find the volume enclosed between the two surfaces z = x 2 + y 2 and z = 8 x 2 y 2. 4. Change of Variables in ouble Integrals A transformation T from the uv-plane to the xy-plane is a function that maps points in the uv-plane to points in the xy-plane, so that T (u, v) = (x, y), where x = g(u, v) and y = h(u, v), for some functions g and h. We consider changes of variables in double integrals as defined by a transformation T from a region S in the uv-plane onto a region in the xy-plane Example 4.1. Let be the region bounded by the straight lines y = 2x + 3, y = 2x + 1, y = 5 x and y = 2 x. Find a transformation T mapping a region S in the uv-plane onto, where S is a rectangular region, with sides parallel to the u- and v-axes.

MULTIPLE INTEGALS 7 The determinant x u y u is referred to as the Jacobian of the transformation T and is written using the notation x v y v (x, y) (u, v). Theorem 4.1 (Change of Variables in ouble Integrals). Suppose that the region S in the uvplane is mapped onto the region in the xy-plane by the one-to-one transformation T defined by x = g(u, v) and y = h(u, v), where g and h have continuous first partial derivatives on S. If (x, y) f is continuous on and the Jacobian is nonzero on S, then (u, v) f(x, y)da = f(g(u, v), h(u, v)) (x, y) (u, v) dudv. S emark. Note that (x, y) (u, v) stands for the absolute value of the Jacobian. emark. Sometimes, it is useful to use the formula (u, v) (x, y) = 1 (x, y). (u, v) Example 4.2. (1) Evaluate (x 2 +2xy) dx dy where is the region bounded by the lines y = 2x + 3, y = 2x + 1, y = 5 x, and y = 2 x. e x y (2) Evaluate the double integral da, where is the rectangle bounded by the lines x + y y = x, y = x + 5, y = 2 x and y = 4 x. (3) Find the area of the region in the first quadrant bounded by the curves xy = 1, xy = 4, y = x, and y = 2x.

8 MULTIPLE INTEGALS (4) Evaluate y 2 = 3x. (5) Evaluate x 2 y 4 dx dy where is the region bounded by xy = 2, xy = 4, y2 = x, and xy dx dy where is the region bounded by the lines x+3y = 1, x+3y = 3, x y = 1, and x y = 2. (6) Evaluate (x 2 +y 2 ) da where is the region bounded by the lines x+y = 1, x+y = 2, 3x + 4y = 5, and 3x + 4y = 6. (7) Find the area of the region bounded by the curves y = x 2, y = 2x 2, x = 3y 2, and x = y 2. (8) Evaluate (x 2 + y 2 ) da where is the region in the first quadrant bounded by y =, y = x, xy = 1, x 2 y 2 = 1. 5. ouble Integrals in Polar Coordinates Let { x = r cos θ T : y = r sin θ be a transformation from rθ-plane to xy-plane. Which transforms regions expressed in polar coordinates to regions in Cartesian coordinates. Example 5.1 (A Transformation Involving Polar Coordinates). Let be the region inside the circle x 2 + y 2 = 9 and outside the circle x 2 + y 2 = 4 and lying in the first quadrant between the lines y = and y = x. Find a transformation T from a rectangular region S in the rθ-plane to the region. Example 5.2. erive the evaluation formula for polar coordinates (r > ): f(x, y)da = f(r cos θ, r sin θ)rdrdθ. Theorem 5.1. Suppose that f(r, θ) is continuous on the region S = {(r, θ) α θ β and g 1 (θ) r g 2 (θ)}, where g 1 (θ) g 2 (θ) for all θ in [α, β]. Then, ˆ β f(r, θ)da = Example 5.3. Evaluate α ˆ g2 (θ) g 1 (θ) f(r, θ)rdrdθ. x2 + y 2 dx dy where is the disk x 2 + y 2 9. Example 5.4. Evaluate dx dy x 2 + y 2 + 1 where is the half disk x2 + y 2 1, y.

Example 5.5. Evaluate MULTIPLE INTEGALS 9 dx dy x2 + y 2 where is the annulus 1 x2 + y 2 4. Example 5.6. Find the area inside the curve defined by r = 2 2 sin θ. Example 5.7. Evaluate origin. (x 2 + y 2 + 3)dA, where is the circle of radius 2 centered at the Example 5.8. Find the volume inside the paraboloid z = 9 x 2 y 2, outside the cylinder x 2 + y 2 = 4 and above the xy-plane. Example 5.9. Evaluate the iterated integral ˆ 1 ˆ 1 x 2 1 x 2 (x 2 + y 2 ) 2 dydx. Example 5.1. Find the volume of the solid bounded by z = 8 x 2 y 2 and z = x 2 + y 2. Example 5.11. Find the volume cut out of the sphere x 2 + y 2 + z 2 x 2 + y 2 = 2y. = 4 by the cylinder

1 MULTIPLE INTEGALS Example 5.12. Find the volume of the solid in the first octant bounded by the circular paraboloid z = x 2 + y 2, the cylinder x 2 + y 2 = 4 and the coordinate planes. Example 5.13. Find the volume of the solid under the surface z = x 2 + y 2 above the xy-plane and inside the cylinder x 2 + y 2 = 2y. Example 5.14. Find the volume V of the solid region T where T is bounded by xy-plane, the cylinder x 2 + y 2 = x and the cone z = x 2 + y 2. Example 5.15. Find the volume V of the solid region T where T is bounded above by the plane z = 1 and below by the cone z = x 2 + y 2. Example 5.16. Evaluate ˆ 1 ˆ 1 x 2 x 2 x 2 + y 2 dy dx. 6. Triple Integrals Suppose that f(x, y, z) is continuous on the box defined by = {(x, y, z) a x b, c y d and r z s}. Then, we can write the triple integral over as a triple iterated integral: f(x, y, z)dv = ˆ s ˆ d ˆ b r c a f(x, y, z)dxdydz. Example 6.1. Evaluate the triple integral 2xe y sin zdv, where is the rectangle defined by = {(x, y, z) 1 x 2, y 1 and z π}. In general, if F (x, y, z) is a function defined on a closed bounded region in space, we evaluate the triple integral by applying a three dimensional version of iterated integration. For instance, if the region can be written in the form = {(x, y, z) (x, y) and g 1 (x, y) z g 2 (x, y)},

MULTIPLE INTEGALS 11 where is some region in the xy-plane and where g 1 (x, y) g 2 (x, y) for all (x, y) in, then Example 6.2. Evaluate y =, z = and 2x + y + z = 4. f(x, y, z)dv = ˆ g2 (x,y) g 1 (x,y) f(x, y, z)dzda. 6xydV, where is the tetrahedron bounded by the planes x =, Example 6.3. Evaluate 6xydV, where is the tetrahedron bounded by the planes x =, y =, z = and 2x + y + z = 4 as in the previous example, but this time, integrate first with respect to x. Example 6.4. Evaluate ˆ 4 ˆ 4 ˆ y x 6 1 + 48z z 3 dzdydx. ecall that for double integrals, we had mentioned that 1dA gives the area of the region. Similarly for solid region in space V = 1dV where V is the volume of the solid. Example 6.5. Find the volume of the solid bounded by the graphs of z = 4 y 2, x + z = 4, x = and z =.

12 MULTIPLE INTEGALS Example 6.6. Find the volume of the solid region bounded by the planes x =, x = 3, z = 1 and z = y 2. Steps to evaluate a triple by integration over a solid region first with respect to z, then with respect to y, and finally with respect to x. Step 1. Sketch: Sketch the region along with its shadow (vertical projection) in the xy-plane. Label the upper and lower bounding surfaces of and the upper and lower bounding curves of. Step 2. Find the z- limits of integration: raw a line M passing through a typical point (x, y) in parallel to the z- axis. As z increases, M enters at z = f 1 (x, y) and leaves at z = f 2 (x, y). These are the z- limits of integration. Step 3. Find the y- limits of integration: raw a line L through (x, y) parallel to the y- axis. As y increases, L enters at y = g 1 (x) and leaves at y = g 2 (x). These are the y- limits of integration. Step 4. Find the x- limits of integration: Choose x- limits that include all lines through parallel to the y- axis. (x = a, x = b). Then the integral is F (x, y, z)dv = ˆ x=b ˆ y=g2 (x) ˆ z=f2 (x,y) x=a y=g 1 (x) z=f 1 (x,y) Follow a similar procedure if you change the order of integration. F (x, y, z) dz dy dx. Change of Variables in Triple Integrals. For a transformation T from a region S of uvwspace onto a region in xyz-space, defined by x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w), (x, y, z) the Jacobian of the transformation is the determinant defined by (u, v, w) x x x (x, y, z) (u, v, w) = u v w y y z u v w z z z u v w Theorem 6.1. Suppose that the region S in uvw-space is mapped onto the region in xyz-space by the one-to-one transformation T defined by x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w), where g, h and l have continuous first partial derivatives in S. If f is continuous in and the

Jacobian (x, y, z) (u, v, w) is nonzero in S, then f(x, y, z)dv = S MULTIPLE INTEGALS 13 f(g(u, v, w), h(u, v, w), l(u, v, w)) (x, y, z) (u, v, w) dudvdw. 7. Cylindrical Coordinates Example 7.1. erive the evaluation formula for triple integrals in spherical coordinates: f(x, y, z)dv = f(r cos θ, r sin θ, z)r dz dr dθ. Example 7.2. Evaluate S e x2 +y 2 dv, where is the solid bounded by the cylinder x 2 +y 2 = 9, the xy-plane and the plane z = 5. Example 7.3. Write f(r, θ, z)dv as a triple iterated integral in cylindrical coordinates if = {(x, y, z) x 2 + y 2 z 18 x 2 y 2 }.

14 MULTIPLE INTEGALS Example 7.4. Evaluate the triple iterated integral ˆ 1 ˆ 1 x ˆ 2 2 x 2 y 2 1 1 x 2 x 2 +y 2 (x 2 + y 2 ) 3 2 dzdydz. Example 7.5. Use a triple integral to find the volume of the solid bounded by the graph of y = 4 x 2 z 2 and the xz-plane. Example 7.6. Find the volume V of where is the solid region bounded by z = x 2 + y 2 and z = x 2 + y 2. ˆ 2 ˆ 2x x ˆ 2 3 Example 7.7. Evaluate z x 2 + y 2 dz dy dx using cylindrical coordinates. Example 7.8. Evaluate (x 2 + y 2 + 1) dz dy dx where is the solid region bounded by the cylinder x 2 + y 2 = 2y and the panes z = and z = 2. Example 7.9. Find the volume V of the solid region bounded by the plane x + z = 1 and the paraboloid z = 1 + x 2 + y 2. Example 7.1. Evaluate zy dv where is the solid region bounded above by the plane z = 1 and below by the cone z = x 2 + y 2. Example 7.11. Evaluate xy dv where is the solid region in the first octant bounded above by the the hemisphere z = 4 x 2 y 2 coordinate planes. and on the sides and the bottom by the

Example 7.12. Evaluate MULTIPLE INTEGALS 15 y 2 z dx dy dz where is the solid region bounded above by the the sphere x 2 + y 2 + z 2 = 4 and below by the cone z = x 2 + y 2. Example 7.13. Evaluate (x 2 + y 2 ) dv where is the solid region between the cylinders x 2 + y 2 = 1 and x 2 + y 2 = 4 and the planes z = 1 and z = 2 in the first octant. Example 7.14. Evaluate ˆ 1 ˆ 1 x ˆ 2 1 1 x 2 x 2 +y 2 x dz dy dx. 8. Spherical Coordinates Example 8.1. erive the evaluation formula for triple integrals in spherical coordinates: f(x, y, z)dv = f(ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)ρ 2 sin φ dρ dφ dθ. S Example 8.2. Find the volume of the sphere of radius a. Example 8.3. Evaluate the triple integral cos(x 2 + y 2 + z 2 ) 3 2 dv, where is the unit ball: x 2 + y 2 + z 2 1. Example 8.4. Find the volume lying inside the sphere x 2 + y 2 + z 2 = 2z and inside the cone z 2 = x 2 + y 2. Example 8.5. Evaluate the triple iterated integral ˆ 2 ˆ 4 x ˆ 4 x 2 2 y 2 2 (x 2 +y 2 +z 2 )dzdydx.

16 MULTIPLE INTEGALS Example 8.6. Evaluate x2 + y 2 + z 2 dx dy dz, where is the ball x 2 + y 2 + z 2 a 2. Example 8.7. Evaluate xz dx dy dz, where is the part of the spherical shell 1 x 2 + y 2 + z 2 4 in the first octant.