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am3 - PRINT NAME SOLUTIONS Calculus IV[2443002 Midterm III...

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PRINT NAME: SOLUTIONS Calculus IV [2443–002] Midterm III 1 Q1 [15 points] 1.1 Part 1 Write down the change of variables formula for triple integrals. 1.2 Answer to part 1 Suppose the change of variables ( x ( u, v, w ) , y ( u, v, w ) , z ( u, v, w )) takes a region S in uvw -space to a region R in xyz -space. Then Z Z Z R f ( x, y, z ) dV = Z Z Z S f ( x ( u, v, w ) , y ( u, v, w ) , z ( u, v, w )) ( x, y, z ) ( u, v, w ) du dv dw where ( x, y, z ) ( u, v, w ) = x u x v x w y u y v y w z u z v z w 1.3 Part 2 Use the change of variables formula to evaluate the volume of the ellipsoid bounded by x 2 a 2 + y 2 b 2 + z 2 c 2 = 1 Show all the steps of your work clearly. You may use the fact that the volume of a sphere of radius r is equal to 4 3 πr 3 . 1.4 Answer to part 2 Recall that Volume = Z Z Z ellipsoid dV Let x = au , y = bv , and z = cw . Then the ellipsoid above becomes a unit ball bounded by the sphere u 2 + v 2 + w 2 = 1 in uvw -space. We also have x u = a , x v = x w = 0, y v = a , y u = y w = 0, and z w = a , z u = z v = 0. Thus we get ( x, y, z ) ( u, v, w ) = a 0 0 0 b 0 0 0 c = abc and, substituting into the change of variables formula yields Z Z Z ellipsoid dV = Z Z Z unit ball abc du dv dw = abc (Volume of unit ball) = 4 π 3 abc .
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2 Q2 [15 points] 2.1 Part 1 Write down the expression for volume element dV in spherical coordinates. [Recall dV = dxdydz in Cartesian coordinates] 2.2 Answer to part 1 The volume element in spherical coordinates is just dV = ρ 2 sin φ dρ dθ dφ 2.3 Part 2 Use spherical coordinates to compute the volume of the solid which lies below the sphere x 2 + y 2 + z 2 = 9 and above the (upper half ( z > 0) of the) cone z 2 = 3( x 2 + y 2 ).
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