Chapter 13 - Multiple Integration (L A TEX) + y ) dy 27. Area of D = D

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1 Andrew Rosen Chapter 13 - Multiple Integration (L A TEX) ouble Integration over Rectangular Regions Fubini s Theorem: If f is continuous on a rectangle with [a, b] [c, d] =, then f(x, y) da = b d a c f(x, y) dy dx = d b c a f(x, y) dx dy Example: (Note - if it s x y + 6 dy, then it becomes xy y + 6y + C) 1 (9x + y) dx dy ( 9 x + xy ) 1 dy ( 7 + y ) dy 7 y + y 5 The average value of an integrable function f over a region is: f(x, y) da f(x, y) da f = Area of = da Remember: If you are looking for the bounds on an integral with respect to dx, the lower bound is the lower function and the upper bound is the higher function. If you are looking for the bounds on an integral with respect to dy, the lower bound is the left-most function and the upper bound is the right-most function, assuming that this is on a typical xy plane ouble Integrals over General Regions With integrals over nonrectangular regions, the order of integration cannot be simply switched. A correct statement would be the following where the bounds are x 1 = a, x = b, y 1 = c, and y = d: f(x, y) da = f(x, y) da = How to find the bounds of a double integral 1 : b h(x) a g(x) d h(y) c g(y) f(x, y) dy dx f(x, y) dx dy 1. etermine with which variable the inner integral is with respect to: dx or dy. The bounding curves determine the limits of integration for the variable determined in the first step 3. The bounds of the remaining variable is the projection of the region on that axis 1 It is important to be able to identify the standard quadric surfaces 1

2 Example: ouble integral with dx on the inside and dy on the outside Compute the iterated integral of the region bounded by y = 1 x, y = 3, and x = with dx on the inside and dy on the outside Step 1: Plot the region Step : etermine the bounds for the dx integral x y + 1 Step 3: etermine the bounds for the dy integral 3 y 1 Step : Write the iterated integral: 1 y+1 3 f(x, y) dx dy Example: ouble integral with dy on the inside and dx on the outside Compute the iterated integral of the region bounded by y = 1 x, y = 3, and x = with dy on the inside and dx on the outside Step 1: Plot the region (see previous plot) Step : etermine the bounds for the dy integral 3 y 1 x Step 3: etermine the bounds for the dx integral x 11 Step : Write the iterated integral: 11 1 x 3 f(x, y) dy dx The volume between two surfaces where g(x, y) f(x, y) is: V = (g(x, y) f(x, y)) da If is a region in the xy plane then the area of that region is: Area = da

3 ouble Integrals in Polar Coordinates Before doing double integration in polar coordinates, it is essential to recall the following identities: r = x + y r = x + y x = r cos(θ), y = r sin(θ) As long as β α π while α θ β and a r b, then the following is the double integral over a polar rectangular region for a function, f(x, y): f(r, θ) da = θ=β r=b θ=α r=a f [r cos(θ), r sin(θ)] r dr dθ With double integrals over non-rectangular polar regions, the order of integration cannot be switched directly. A more general expression for the double integral over a general polar rectangular region where g(θ) r h(θ) and β α π with α θ β for a function, f(x, y), is: f(r, θ) da = θ=β r=h(θ) θ=α r=g(θ) f [r cos(θ), r sin(θ)] r dr dθ Note: on t forget to multiply the integrand by a factor of r! Area of Polar Regions: A = da = θ=β r=h(θ) θ=a r=g(θ) r dr dθ The following trigonometric identities are crucial to integration of squared trigonometric functions: sin (θ) = 1 (1 cos(θ)) cos (θ) = 1 (1 + cos(θ)) It is essential to be able to recall the typical trigonometric values for sine, cosine, and tangent 3

4 Example: Set up an equation to find the area of 1 leaf of the rose, r = cos(θ) 1) Plot the function: ) Write the bounds r cos(θ) and π θ π Note: When writing the bounds of integration for θ, make sure that it is going from a lower value to a higher value and make sure it is the correct region of the boundary. In this case, π to 7π would be incorrect since it d sweep more than one leaf. 3) Set up double integral A = π π cos(θ) r dr dθ Triple Integrals f(x, y, z) dv = b h(x) H(x,y) a g(x) G(x,y) f(x, y, z) dz dy dx Note: Five other orders of integration could be set up based on Fubini s Theorem How to Find the Bounds of Integration: 1. Your first set of limits can be figured out if you imagine traveling through the three-dimensional boundary in the direction of the axis of the variable you re integrating. Find where you enter the three-dimensional boundary for the first time and where you exits. Write this as an inequality (eg: a z b, where the integral is with respect to dz).. Set the first variable of integration to zero (so, if you integrated with respect to z first, make z = ) to create a plane. It might even be helpful to draw a new graph in a normal -variable Cartesian plane. Now travel in the direction of the middle variable of integration. Find the entrance and exit sites. 3. Now you are left with one variable to find the limits of integration for. The bounds of the remaining variable is the projection of the region on that axis.

5 How to switch the bounds of a triple integral (and how to graph a 3 function given the bounds): 1. Since the middle and outer integrals bounds are the projection of the 3 region, plot this projection first. Be careful of this major fact: If you have dx dy dz, you d graph a yz projection first. If one of the bounds for, let s say, y is something like y = z then you can t graph that directly! See what z equals and substitute that in for y = z.. From here, the bounds for the middle integral can be found as usual. The bounds for the outer integral are the projection of the graph on the axis of the variable that the outer integral is with respect to (make sure that the outer integral has constants for the bounds) 3. Now, extrude this projection in the dimension of the inner integral variable. The bounds for the inner integral can then be found as normal.. The key is to realize which bounds are functions of which variables and to adjust them accordingly. 5

6 Example: Set up a triple integral to find the volume of the given solid region in the first octant bounded by the plane 1x + 16y + 1z = 8 and the coordinate planes. Given graph of the boundary: Step 1: Pick an order of integration. I will pick dz dy dx arbitrarily. Any order can be used. Step : Imagine going through this boundary in the z axis. It enters at z = and exits at the equation of the plane solved for z, which is x y 3. Therefore, z x y 3 Step 3: Set z = to create a new Cartesian plane. Re-plotting the boundary might be helpful. The graph is shown below. Step : Find the boundaries for y. The hypothetical knife would enter at y =. Now you need the exit point. There are two easy ways to do this. One is to simply solve for y when z =. This would make y = 3 x + 3. Another way to do this, which is equally as easy, is to recognize that the slope of a line is the y x, which is 3. This, in conjunction with the knowledge that the y intercept is (, 3) can yield the equation of the line, which is 3 x + 3. Therefore, y 3 x + 3. Step 5: Find the remaining boundary for x. The projection of the previous boundary on the x axis yields x. Step 6: Use this information to piece together a triple integral. It would thus be: 3x +3 x y 3 dx dy dz 6

7 13.5, Part I - Triple Integrals in Cylindrical Coordinates The triple integral of f over in cylindrical coordinates is: f(x, y, z) dv = β h(θ) H(r cos(θ),r sin(θ)) α g(θ) G(r cos(θ),r sin(θ)) f(r, θ, z) dz r dr dθ The following are important conversion rules between rectangular and cylindrical coordinates: 1. x = r cos(θ). y = r sin(θ) 3. z = z Example: Set up, but do not evaluate, x x x +y (x + y ) dz dy dx in cylindrical coordinates. Step 1: Plot the boundary region. The tip of the cone is at (,, ) and opens from z = to z = Step : Find the boundaries of z by converting to cylindrical coordinates. The boundaries become r z. Also note that the integrand itself becomes r. Step 3: Collapse the cone so that z = and plot it (indicated by shaded region) Remember to check the boundaries given Step : Find the boundaries for r and θ. With this graph, it is clear that r. The angle must sweep from a smaller angle to a higher angle for the boundary, so π θ π. Step 5: Set up the triple integral. It would become, with the boundaries already determined: π π r r r dz dr dθ 13.5, Part II - Triple Integrals in Spherical Coordinates There are three new coordinates to recognize in spherical coordinates: 1. ρ is the distance from the origin to a point, P. φ is the angle between the positive z-axis and an arbitrary line OP that goes from to π 3 3. θ is the same angle as in cylindrical coordinates and measures rotation about the z-axis relative to the positive x-axis 3 φ = is a line from the origin going upward vertically and φ = π is a line from the origin going downward vertically 7

8 The triple integral of f over a region,, in spherical coordinates is: f(ρ, φ, θ) dv = β b h(φ,θ) α a g(φ,θ) f(ρ, φ, θ)ρ sin φ dρ dφ dθ The following are important conversion rules between rectangular and spherical coordinates: 1. x = ρ sin φ cos θ. y = ρ sin φ sin θ 3. z = ρ cos φ Also note: ρ = r + z and ρ = x + y + z = r + z Example: Set up, but do not evaluate, the following integral in spherical coordinates: is a sphere of radius 6. e (x +y +z ) 3 Step 1: Convert known functions to spherical coordinates. The triple integral becomes Step : Find the bounds for ρ. This will extend from to the sphere, which is 6 ( e r) 3 dv where Step 3: Find the bounds for φ. The bounds for this will be from to π. The reason that it is not from to π is because the bounds of θ take care of that part of the sweeping action Step : Find the bounds for θ. The bounds for this will be one full revolution from to π Step 5: Set up the triple integral. It will be: π π 6 e 8ρ3 ρ sin φ dρ dφ dθ dv Example: Set up, but do not evaluate, an expression for the volume of the smaller region cut from the solid sphere ρ 1 by the plane z = 7. Step 1: Find the bounds for ρ. An equation for a plane is given by a sec φ because z = 7 is the plane and z = ρ cos φ, so the lower bound for ρ is 6 sec φ, and the upper bound is 1. Step : Find the bounds for φ. The lower limit is φ =, and the upper limit of π can be found by setting 3 7 equal to 1 and solving for φ. cos φ Step 3: Find the bounds for θ. These are simply those of a circle in order to create one full revolution. Therefore, the lower bound is and upper bound is π. Step : Write out the triple integral. π π/3 1 7/ cos φ ρ sin φ dρ dφ dθ 8

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