Question

Triple Integration Problem using cylindrical coordinates:

Answer 1

\[\int\limits_{0}^{2}\int\limits_{0}^{\sqrt{4-y^{2}}}\int\limits_{\sqrt{x ^{2}+y ^{2}}}^{\sqrt{8-x ^{2}-y ^{2}}}z^{2}dzdxdy\]

Answer 2

I don't want to see the problem done; I just don't know how to convert the limits of integration to suit Cylindrical and I can't find anything helpful. Can anyone show me how I would go about it; or a resource on how?

Answer 3

@dan815

Can you help when you see this?

Answer 4

I will post what work I try to do below.

Answer 5

so for the starters for the dz integral the lower bound is just going to be r.. for the top i decided to plugin for x and y where x = rcos(theta) and y = rsin(theta)

Answer 6

simplifying that down you would find that upper bound is:
\[\sqrt{8 - r ^{2}}\]

Answer 7

Not sure if im on the right track or not though for what i'm doing..

Answer 8

for the dx portion we have the upper half of a circle with a radius of 2 so would the limit of integration for dx be from 0 to 2?

Answer 9

Graphing for x = \[\sqrt{4-x ^{2}}\] I solved that for y = and then graphed and found that the theta values are ranging from 0 to pi

Answer 10

Not even remotely sure if i'm right on what to do..

Answer 11

This was my thoughts on what the integrals would look like:\[\int\limits_{0}^{\pi}\int\limits_{0}^{2}\int\limits_{r}^{\sqrt{8-r ^{2}}}z ^{2}rdzdrd \theta\]

Please correct me if i'm wrong.

Answer 12

Also unsure about Z since Z = Z when converting from rectangular to cylindrical. I figured that z^2 would stay the same. and once you integrate it you'd have r's where z's are.

Answer 13

@FutureMathProfessor can you look at this question I asked but nobody answered so I just closed it?

Answer 14

Our Y coordinate goes from 0 to 2 so those essentially straight lines