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Unformatted text preview: Math 192 Final Exam Solutions Spring 2007 1 z y x 1. (a) Write down a triple integral for the volume of the solid in the first octant bounded by the planes x + y + z = 2 and z = 1. Solution : In order to do this without breaking the region up into several regions the only way is to do the integration in the order dx dy dz or dy dx dz . For dx dy dz the shadow of the region is its quadrilateral face in the yzplane, so the integral is Z 1 Z 2 z Z 2 y z dx dy dz (b) Compute the volume using the triple integral. Solution : Z 1 Z 2 z Z 2 y z dx dy dz = Z 1 Z 2 z (2 y z ) dy dz = Z 1 (2 y y 2 2 yz ) fl fl fl 2 z dz = Z 1 (2 2 z + z 2 2 ) dz = (2 z z 2 + z 3 6 ) fl fl fl 1 = 7 6 2. Evaluate the integral Z 1 Z 1 y 1 / 3 1 x 4 + 1 dx dy . Solution : To carry out the integration we have to switch the order of integration. This gives Z 1 Z x 3 1 x 4 + 1 dy dx = Z 1 x 3 x 4 + 1 dx = 1 4 log( x 4 + 1) fl fl fl 1 = 1 4 log 2 3. Evaluate the integral Z 1 Z √ 1 x 2 Z 1+ √ 1 x 2 y 2 ( x 2 + y 2 ) 1 / 4 2 z tan 1 ( y/x ) dz dy dx . Bonus: Sketch the region of integration. Solution : Cylindrical coordinates work well here, especially since tan 1 ( y/x ) = θ . The shadow of the 3dimensional region of integration in the xyplane is determined by the limits of integration in the first two integrals Z 1 Z √ 1 x 2 , so the shadow is the quarter of the unit disk x 2 + y 2 ≤ 1 in the first quadrant. Then in cylindrical coordinates we get Z π/ 2 Z 1 Z 1+ √ 1 r 2 r 1 / 2 2 z θ r dz dr dθ = Z π/ 2 Z 1 z 2 θr fl fl fl fl 1+ √ 1 r 2 r 1 / 2 dr dθ = Z π/ 2 Z 1 θr (1 + 2 p 1 r 2 + 1 r 2 r ) dr dθ = Z π/ 2 Z 1 θ (2 r + 2 r p 1 r 2 r 2 r 3 ) dr dθ = Z π/ 2 θ ‡ r 2 2 3 (1 r 2 ) 3 / 2 r 3 3 r 4 4 · fl fl fl fl 1 dθ = Z π/ 2 ‡ 13 12 · θ dθ = 13 12 θ 2 2 fl fl fl fl π/ 2 = 13 12 π 2 8 Math 192 Final Exam Solutions Spring 2007 Here is what the region of integration looks like. It is one quarter of a solid of revolution. z y x 4. (a) Consider a rectangular solid resting on the xyplane with faces parallel to the coordinate planes and with its upper corners on the sphere x 2 + y 2 + z 2 = 1. If the corner in the first octant is at the point ( a, b, c ) and the solid is made from a material of variable density δ ( x, y, z ) = z , show that the mass of the solid is M ( a, b, c ) = 2 abc 2 ....
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This note was uploaded on 06/01/2008 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 PANTANO
 Math

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