Question

Use cylindrical coordinates to evaluate the triple integral
∫∫∫sqrt(x2+y2)dV where E is the solid bounded by the circular parabaloid
z=9−16(x2+y2) and the xy plane.

Answer 1

This was the integral that I set up:

\[\int\limits_{0}^{2\pi}\int\limits_{0}^{3/4}\int\limits_{0}^{9-16r^2}rdzdrd \theta\]

And I was not able to wind up with the correct answer.

Answer 2

you do z first

Answer 3

you integrate and get z b/c u only have a dz

Answer 4

so u get 9-16r^{2} rdrdtheta

Answer 5

I know I just think that my second limit of integration might be wrong. I know how to evaluate the integral.

Answer 6

then for r you get 9r-16r^3 drdtheta

Answer 7

\[\int\limits_{0}^{3/4} 9r-16r^{3} drd\]

Answer 8

\[\frac{ 9r^2 }{ 2} - 4r^4 \]

Answer 9

from 0 to 3/4

Answer 10

no there is a function r, so it should be r^2.
then int[9r2-16r4]0-3/4 *2pi

Answer 11

r3-16/5r5]0-3/4 * 2pi is the final answer.