QUIZ 4 (CHAPTER 17) SOLUTIONS MATH 5 FALL 8 KUNIYUKI SCORED OUT OF 15 POINTS MULTIPLIED BY.84 15% POSSIBLE 1) Reverse the order of integration, and evaluate the resulting double integral: 16 y dx dy. Give a simplified exact answer; do not approximate. 4 y 113 + x 9 Sketch the region of integration; you may use different scales for the x- and y- axes. ( points) Sketch the region of integration, R. It may help to label the limits of integration: Let I y 16 y x x 4 y y 113+ x 9 dx dy We aim / fix y, shoot x, and slide y. Reverse the order of integration. Solve x 4 y for y: y x 4 ( x )
We aim / fix x, shoot y, and slide x. I x x x x x x y x 4 y y x 4 y y 113 + x 9 dy dx 1 y dy dx 113 + x 9 1 y 113 + x 9 1 113 + x 9 x 8 113 + x 9 dx ( x ) 4 y x 4 y dx dx We now perform a standard u-substitution. Let u 113 + x 9 du 9x 8 dx x 8 dx 1 9 du You can also Compensate : x 8 dx 1 9x 8 113+ x 9 9 dx 113+ x 9 Change the limits of integration: x u 113 + ( ) 9 u 113 x u 113 + ( ) 9 u 65
1 9 1 18 1 18 u 65 u 113 65 113 u 1/ 1/ 1 18 u du u u 1/ du 65 65 113 65 113 1 u 9 113 1 ( 65 113) 9 5 113 9 ( Note: This is about 1.59665. ) ) Evaluate e 3x +3y dy dx. Your answer must be exact and simplified. 4 x You do not have to approximate it. (16 points) Let s call this double integral I. It may help to label the limits of integration: I x x y 4 x y e 3x +3y dy dx Sketch the region of integration, R. If we aim / fix any x in,, then y is shot from y to y 4 x. The graph of y 4 x is a semicircle, the upper half of the circle of radius centered at the origin. Since the x-interval is,, we do graph the entire semicircle. Observe that the integrand, e 3x +3y, is continuous throughout R.
Convert to polar coordinates. Why? R is well suited for that. The integrand is well suited for that. It looks like using Cartesian coordinates will be difficult. I x x y 4 x y ( ) da e 3 x + y R e 3r da R r e 3x +3y dy dx e 3r rdrd At this point, we can separate the double integral into a product of single integrals. This is because the limits of integration are all constants, and the integrand is separable. d r re 3r dr Evaluate the first integral: d To evaluate the second integral, u-substitution: r re 3r dr, we perform a standard Let u 3r du 6rdr rdr 1 6 du You can also Compensate : r re 3r dr 1 6 Change the limits of integration: u 3 ( ) u r u 3 ( ) u 1 r 6re 3r dr
I 6 1 6 1 6 eu u 1 u e u du 1 ( 6 e1 e ) e u du ( 6 e1 1) ( 85,18) 3) Find the surface area of the portion of the graph of z 7x + 1 y that lies over R, where R is the region in the xy-plane bounded by the rectangle with vertices (, ), ( 4, ), (, ), and ( 4, ). Give an exact answer; you do not have to approximate it. Distance is measured in meters. Major Hint: Use the following Table of Integrals formula: ( points) a + u du u a + u + a ln u + a + u + C Let S represent the desired surface area. S 1+ f x x, y R [ ( )] + f y ( x, y) [ ] da Let f ( x, y) 7x + 1 y f x, y x ( ) 7, and f y ( x, y) y Here s R :
x 4 y S 1+ 7 + y dy dx x x 4 y y 1+ 49 + y dy dx x x 4 y y 5 + y dy dx x y At this point, we can separate the double integral into a product of single integrals. Reversing the order of integration would have been easy but unhelpful. 5 + y 4 dy dx 5 + y dy 4 We re going to need a Table of Integrals formula or a trig sub (Section 9.3 in Swokowski s Classic Edition): y 5 tan, or y 5 tan. We use a 5 and u y in the given Table of Integrals formula. 4 u a + u + a ln u + a + u 4 y 5 + y + 5 ln y + 5 + y y 5 + y + 5ln y + 5 + y 5+ 54 + 5ln + 54 ( ) + 5ln + 5 + ( ) 5+ ( ) + 5ln 5 ( ) ( ) + 5ln( + 3 6 ) + 5ln( 5 ) ( ) 3 6 ( ) + 5ln + 5 + ( ) 1 6 + 1ln + 3 6 4 3 6 + 5ln + 3 6 ( ) 1ln 5 ( ) 5ln( 5 ) ( ), or ( ), or 4 3 6 + 5ln + 3 6 5, or + 3 3 4 3 6 + 5ln square meters 5 Note: This is about 57.31399 square meters.
4) Set up a triple integral for the volume of the solid bounded by the graphs of y x 3x, x y, z 5, and z e x e y 5. Make sure your triple integral is as detailed as possible; do not leave in generic notation like R or f. Also, do not leave dv in your final answer; break it down into d ( variable)d ( variable)d ( variable). Do not evaluate. (15 points) Let R, the region of integration, be the projection of this solid on the xy-plane. We will sketch R in the xy-plane. The graphs of y x 3x (the parabola in blue) and x y, or equivalently y x, (the line in red) are drawn below. If we were to include z-coordinates and look at the 3-D picture, the graph of y x 3x would be a parabolic cylinder, and the graph of y x would be a plane.
Find the x-coordinates of the intersection points: Solve the system y x 3x y x for x. x 3x x x 5x x( x 5) x or x 5 Because the blue parabola opens up, it makes sense that it lies below the red line on the x-interval, 5. You could also use the test value method to check this out for a value of x strictly between and 5, say 1. (Observe that x 3x and x represent continuous functions of x.) Test x 1 y x 3x 1 x1 () 31 () parabola on bottom Test x 1 y x x1 1 () line on top ( ) ( ) Observe: z e x e y 5 z 5 + e x + e y. Because e x > and e y > throughout R, it must be true that z > 5 throughout R, as well. The graph of z 5 + e x + e y will be the top graph relative to the bottom graph of z 5. If x, y from z 5 up to z 5 + e x + e y. V Q ( ) is fixed in R, then z is shot dv, where Q is the region represented by the solid, so the desired integral is: V x 5 x y x y x 3x z 5+e x +e y z 5 dz dy dx Below is a peek inside the region whose volume we are setting up. The floor is part of the graph of z 5, a horizontal plane. The ceiling is part of the graph of z 5 + e x + e y, but the ceiling rises too quickly for us to effectively capture it all.
5) A solid is bounded by the graph of z x + y, the graph of x + y 16, and the xy-plane. The density at any point in the solid is directly proportional to the square of the distance from the point to the xy-plane. Find the center of mass of the solid; it turns out to be a point outside of the actual solid. Use cylindrical coordinates to complete the problem. If you use any short cuts, explain them precisely! Warning: You may run into some very big numbers along the way! (9 points) The graph of z x + y is a paraboloid opening along the positive z-axis. In cylindrical coordinates, its equation is z r. The graph of x + y 16 is a right circular cylinder of radius 4; the z-axis is its axis. In cylindrical coordinates, its equation is r 16, or. The density function is given by ( x, y,z) kz. Let ( x, y, z) be the center of mass. By symmetry of Q, the region corresponding to the solid, about the planes x (i.e., the yz-plane) and y (i.e., the xz-plane) and the fact ( ) is even in x and y, we have x and y. that x, y,z Find C, the circle of intersection between the paraboloid and the cylinder. z r ( ) z 4 z 16 The projection of C on the xy-plane is the circle C 1 of radius 4 in the xy-plane centered at the origin. C 1 is the boundary of R, the projection of the solid on the xy-plane.
Observe: For a fixed ( x, y) in R, the corresponding z-coordinates shoot from the plane z (i.e., the xy-plane) up to the paraboloid z r. Find the mass of the solid (corresponding to region Q ): m Q ( x, y, z) dv kz dv Q dv z r kz rdzdrd z z r k rz dz dr d z z 3 k r 3 k 3 k 3 k 3 k 3 r z 3 r r z r z z r z ( ) 3 ( ) 3 r r 6 dr d r 7 dr d dr d dr d dr d At this point, we can separate the double integral into a product of single integrals. k r 7 dr 3 k r 8 3 8 ( ) 8 k 4 3 8 k 65536 3 8 ( ) k 3 819 16,384k 3 d mass units
Find z. z M xy m, where: M xy Q z ( x, y, z) dv z kz dv Q dv z r z kz rdzdrd z z r k rz 3 dz dr d z z 4 k r 4 k 4 k 4 k 4 k 4 r z 4 r r z r z z r z ( ) 4 ( ) 4 r r 8 dr d r 9 dr d dr d dr d dr d At this point, we can separate the double integral into a product of single integrals. k r 9 dr 4 k r 1 4 1 k ( 4) 1 4 1 k ( 4) 1 4 5 1 ( 4) 9 k 5 6,144k 5 d distance-mass units
Thus, z M xy m 6,144k 5 16,384k 3 6,144 k 3 5 16,384 k 6,144 3 16,384 5 ( 16) 3 5 48 5 distance units The center of mass is: ( x, y, z),, 48 5, or (,, 9.6) Note 1: Observe that this point actually lies outside of the solid. It lies 6% of the way along the z-axis from the xy-plane to the plane z 16, which contains the top of the solid. Note : Although the shape of the region favors lower z-values, the density function favors higher z-values.
6) Consider the Cartesian (or rectangular) coordinates x, y, and z, and the spherical coordinates,, and. Verify that, for the spherical coordinate ( x, y, z) transformation, the Jacobian (,,) sin. Begin by expressing x, y, and z in terms of,, and ; you may do this from memory without showing work. Work out an appropriate determinant, and clearly show each step; do not simply use a geometric argument. (5 points) The spherical coordinate transformation: x sin cos y sin sin z cos Find the desired Jacobian: ( ) ( ) x, y, z,, x x x y y y z z z ( or the determinant of the transpose) sin cos sin sin cos cos cos cos sin sin sin sin sin cos Let's expand this determinant along the third row so that we can exploit the "." + ( sin sin ) ( sin cos) + sin sin cos sin sin cos cos cos cos sin cos sin ( )( sin sin cos sin ) ( sin cos) ( sin cos cos cos) sin sin
( ) ( sin )( sin + cos ) ( ) ( cos) ( sin + cos ) sin sin sin cos ( ) ( sin )() 1 ( sin cos) ( cos) () 1 sin sin ( )( sin ) ( sin cos) ( cos) sin sin sin sin + sin cos ( sin) ( sin + cos ) ( sin) () 1 sin