# 12 1 dz da π 5sin θ 100 r 2 12 100 r 2 12 1 dzrdrd

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) 1/2 1 dz dA π 0 5sin( θ ) 0 (100 r 2 ) 1/2 (100 r 2 ) 1/2 1 dzrdrd θ . Note: r 5 sin( θ ) r 2 5 r sin( θ ) x 2 y 2 5 y x 2 ( y 5 2 ) 2 5 2 2 . By sketching the rectangular auxiliary r θ -graph, you can see that the circle is traced out as θ runs from 0 to π . ______________________________________________________________________ 4. (10 pts.) Write down a triple iterated integral in cartesian coordinates that would be used to find the volume of the solid G bounded by the surface y = x 2 and the planes y + z = 4 and z = 0, but do not attempt to evaluate the triple iterated integral you have obtained. [Sketching the traces in the coordinate planes will help.] G 1 dV 2 2 4 x 2 4 y 0 1 dzdydx or 4 0 y 1/2 y 1/2 4 y 0 1 dzdxdy . This, of course, was a silly homework problem.

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TEST4/MAC2313 Page 3 of 6 ______________________________________________________________________ 5. (10 pts.) Write down the triple iterated integral in spherical coordinates that would be used to compute the volume of the solid G within the cone defined by φ = π /4 and between the spheres defined by ρ = 4 and ρ = 9 in the first octant . Do not attempt to evaluate the interated integrals. G 1 dV π /2 0 π /4 0 9 4 ρ 2 sin( φ ) d ρ d φ d θ . This is an easy to understand spherical wedge. Draw your own side view in the xz-plane. The only wrinkle is the first octant noise that restricts θ so that 0 θ π /2. ______________________________________________________________________ 6. (10 pts.) Compute the surface area of the portion of the hemisphere defined by z = (16 - x 2 - y 2 ) 1/2 that lies between the planes defined by z = 1 and z = 2. Obviously we have passed to polar coordinates along the way and then used the u-substitution u = 16 - r 2 to accomplish our goal. The intersection of the hemisphere with the two planes are a couple of easy to understand circles. These provide the r limits of integration.
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