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Mat 241 Chapter 15 Test 2010 A & B KEY Name___________________________________ Professor Schultz #1. (5 pts) The shown integral is somewhat “evil” in the order of integration specified. To overcome this obstacle, compute the exact integral by reversing the order of integration. 3 x2 ye ye ye 0 1 x8 dxdy 0 0 x8 dydx 0 3x8 y3 du let u x 2 , xdx 2 8 2 2 x3 2 x2 2 x2 2 x3 0 2 3 2 2 x2 x 3 ex 1xe dx 0 3x8 dx 0 3 dx 4 4 eu eu 1 dx e4 1 6 6 0 6 0 #1. (5 pts) The “sin-log” function has no known antiderivative. To overcome this obstacle, compute the exact integral by reversing the order of integration. Note: log(x) denotes the natural logarithm ln(x) for most mathematicians, engineers, and computer languages. The notation ln is for greenhorns. e sin x 0 y log x dxdy 1 e 1 e log x 0 log( x ) e sin x sin x y dx dydx log x log x 0 1 e sin x log( x) sin x 0 e dx sin x dx cos x 1 log x log x 1 1 cos e cos 1 cos 1 cos e e #2. (10 pts) Using a double or triple integral to find the volume of the solid tetrahedron with vertices: 0,0,0 , 0,0,1 , 0,2,0 , and 2,2,0 . To determine the plane’s equation we do the following: 1. Select a vertex from: 0,0,1 , 0,2,0 ,or 2,2,0 . 2. Create two vectors using the selected point and the other two remaining points. 3. Utilizing those two vectors we compute the cross-product. 4. With the normal vector and one of the three points we determine the plane’s equation upon which we can integrate to get our desired volume. P 0,0,1 ,Q 0,2,0 ,and R 2,2,0 PQ 2,0,0 ;PR 2, 2,1 i PQ PR 2 j k 0 0 0 0 i 2 0 j 4 0 k 0,2,4 2 2 1 The plane's equation is: 0 x - 0 2 y 0 4 z 1 0 2y 4z 4 0 z 1 2 2 V y 2 1 y 2 0 x 0 2 2 2 y y2 x2 1dydx 1 dydx y dx 1 x dx 2 4 x 4 0 x 0 0 2 2 2 x2 x3 x2 8 4 8 2 1 x dx x 2 4 12 2 0 12 2 12 3 0 2 Of course, we could just use the 7th grade formula: Vpyramid 1 1 1 2 area_ of _ the _base height 2 2 1 3 3 2 3 #3. (5 pts) Evaluate the following integral exactly. A perfect candidate for polar coordinates. x2 sin x 2 y 2 dydx 0 sin r2 rdrd let u r2 ; 0 0 du rdr 2 1 1 sin udud cosu 0 d 1 1 d d 20 20 0 0 0 #4. (6 pts) For the pictured solid assume that the density is proportional to the distance to the xz - plane. The solid is formed by the following surfaces: y x, z 1 x2 , x 1, y 2x A. Set – up ONLY! An integral to compute the mass of the solid in the dydzdx order. ALSO sketch the plane projection. 1 1 x2 2x mass kydydzdx 0 2 0 x z 1.8 1.6 1.4 1.2 1 0.8 z 1x2 0.6 0.4 0.2 x -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 -0.2 B. Set – up ONLY! An integral to find sketch the plane projection. M xz in the dzdydx order. ALSO 1 2x 1 x2 Mxz 0 x y ky 2dzdydx 0 4.5 4 3.5 3 2.5 2 1.5 R 1 yx 0.5 x -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 I y in the dydxdz order. ALSO C. Set – up ONLY! An integral to find sketch the plane projection. 1 1z2 2x Iy 0 x 0 2 z2 kydydxdz x 2 z 1.8 1.6 1.4 1.2 1 0.8 z 1x2 0.6 0.4 0.2 x -0.8 -0.6 -0.4 -0.2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 -0.2 #5. (10 pts) FIND THE VOLUME of the solid that is bounded between the two cones given by: z x 2 y2 and z 3 x2 y2 And between the two hemi-spheres: z 4 x 2 y2 and z 1 x2 y2 To help you, the projection in the yz-plane is shown below. The area to be swept into a volume is shown as z z 3x2 y2 z 4 x2 y2 z x2 y2 z 1 x2 y2 y Looks like a case for spherical coordinates to me! I’ll need to find ranges for , ,and . From the picture: zinnersphere 1 x 2 y2 x 2 y 2 zoutsidecone 1 r2 r2 r 1 2 and zoutersphere 4 x 2 y 2 3 x 2 y 2 zinsidecone 4 x2 y2 3 x2 y2 4 r2 3r2 r 1 Our first phi angle is: z z 3x2 y2 z 4 x2 y2 z x2 y2 1 2 z 1x y 2 2 6 y Our second phi angle is: z z 3x2 y2 z 4 x2 y2 z 1 x2 y2 z x2 y2 1/sqrt(2) 4 1 y So we have: 1 2, ,0 2 6 4 V 2 4 2 2 4 0 1 6 6 2 sin ddd 0 3 2 4 7 sin dd sin dd 30 1 3 2 6 2 2 2 7 7 7 2 3 4 cos d cos cos d d 30 3 4 6 3 2 2 6 0 0 7 6 3 2 2 d 73 3 2 0 #5. (10 pts) Calculate the exact volume of the child’s plastic chair formed by the intersection of x2 y2 1 , 2 2 the two surfaces z x 5 and z 4 y the cylinder and having y0 (Note: The chair has been rotated so you can see inside it.) Let x and y equal zero to determine top and bottom surfaces. z x 2 5 and z 4 y 2 when x y 0 z 5 & z 4 We are integrating over the circle x 2 y 2 1 with y 0. 1 V 1 1 x2 4 y2 0 x 5 2 1 4 r2 sin2 1dzdydx rdzdrd 0 0 r cos 5 2 2 1 4 r2 sin2 r2 cos2 5 rdrd 0 0 1 9 r2 sin2 r2 cos2 rdrd 0 0 1 9 r2 rdrd 0 0 1 9r r 3 drd 0 0 1 9r2 r 4 17 17 d d 4 4 0 4 0 2 0