Volume of bounded region

Question

I would like to find the volume of the region bounded by:

$x^2 + y^2 = 1$

I just need help setting up the integral. Any and all help is greatly appreciated. Thank you!

My work thus far:

$V = \int_0^{2 \pi} \int _0 ^1 \int _ {r^2(1+2sin2 \theta)} ^ {4-r^2(1+2sin2 \theta)}rdzdrd \theta$

Welcome to Math.SE! I've rollbacked your [edit]. Please don't deface your own post so that others (e.g. @Shailesh from LQ review) understand your question and your attempts. — GNUSupporter 8964民主女神地下教會, Apr 27 '18 at 00:08

score 2 · Accepted Answer · edited Apr 26 '18 at 22:42

2

First find the intersection of the the two surfaces, to find $$ x^2+y^2 =1.$$

Then set up the integral in cylindrical coordinates.

$$ V= \int _0^{2\pi} \int _0^1 \int _{r^2(1+2\cos ^2 \theta)} ^{4-r^2(1+2\sin ^2 \theta)} dzr dr d \theta $$

edited Apr 26 '18 at 22:42

amWhy

answered Apr 26 '18 at 22:33

For the inner-most integral, how did you get those limits of integration? – Matthew Apr 26 '18 at 22:39
The lower limit is the lower surface in polar form and the upper limit is the upper surface in polar form. $( x=r\cos \theta , y=r\sin \theta)$ – Mohammad Riazi-Kermani Apr 26 '18 at 22:42
$ \sin^2 \theta + \cos ^2 \theta =1$ – Mohammad Riazi-Kermani Apr 26 '18 at 22:59
The correct answer is $2\pi$ and both methods give the same correct answer. – Mohammad Riazi-Kermani Apr 26 '18 at 23:52
Thank you for your help, I need to improve my integration skills - obviously! I hope you have a great weekend. – Matthew Apr 26 '18 at 23:54
Thanks and have a good one. – Mohammad Riazi-Kermani Apr 26 '18 at 23:57

1 Answers1