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Unformatted text preview: Solution to Problem Set #8 1. (20 pt) Find the volume of an ice cream cone bounded by the hemi sphere z = 8 x 2 y 2 and the cone z = x 2 + y 2 . The graphs above are the graphs of z = 8 x 2 y 2 , z = x 2 + y 2 and their intersection. Solution. –4 –2 2 4 x –4 –2 2 4 y –4 –2 2 4 –4 –2 2 4 x –4 –2 2 4 y –4 –2 2 4 MATH 2850: page 1 of 4 Problem Set #8 MATH 2850: page 2 of 4 –4 –2 2 4 x –4 –2 2 4 y 2 4 6 8 10 The region is bounded above by the hemisphere z = 8 x 2 y 2 and below by the cone z = x 2 + y 2 . We have x 2 + y 2 ≤ z ≤ 8 x 2 y 2 . Thus x 2 + y 2 ≤ z 2 ≤ 8 x 2 y 2 and x 2 + y 2 ≤ 4 In polar coordinates, this region x 2 + y 2 ≤ 4 is R = ( r, θ ) : 0 ≤ r ≤ 2 , ≤ θ ≤ 2 π . Note that 8 x 2 y 2 = √ 8 r 2 and x 2 + y 2 = √ r 2 = r . Hence, we can compute the volume of the ice cream cone by finding the volume under the graph of √ 8 r 2 above the disk R = ( r, θ ) : 0 ≤ r ≤ 2 , ≤ θ ≤ 2 π and subtracting the volume under the graph of r above R . Therefore, we have Volume = 2 π 2 ( √...
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This note was uploaded on 10/21/2009 for the course MATH 223 taught by Professor Punosevac during the Spring '08 term at Arizona.
 Spring '08
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