Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a problem with calculating volume of given function over the area.

I enclose an image with my solution, however, I got a bad answer.
According to my book the answer should be pi when I got 2*pi.

Sorry for the quality, I couldn't get better with my phone. Thank you in advance for your help!


share|cite|improve this question
+1 for "what you tried". Maybe writing it down in $\LaTeX$ would have helped you to solve this on your own... – draks ... Aug 26 '13 at 19:00
next time I'll do that, thanks! – VsMaX Aug 26 '13 at 19:10
up vote 0 down vote accepted

There is something wrong in:

$$ \int_0^{2\pi} \int_0^r r(r\cos \phi +1)(r\sin\phi +1) d\phi dr= \int_0^{2\pi} \int_0^r (r^2\cos \phi +\color{red}r)(r\sin\phi +1) d\phi dr=\\ \int_0^{2\pi} \int_0^r (r^3\sin\phi\cos \phi +\color{red}r\sin\phi + r^2\cos \phi +\color{red}r) d\phi dr=\\ \int_0^{2\pi} \Biggr[ \frac14r^4\sin\phi\cos \phi +\frac12\color{red}{r^2}\sin\phi + \frac13r^3\cos \phi +\color{red}{\frac12 r^2}\Biggr]_0^1 d\phi =\\ \int_0^{2\pi} \frac14\sin\phi\cos \phi +\frac12\sin\phi + \frac13\cos \phi +\color{red}{\frac12 } d\phi = \dots=\pi $$

share|cite|improve this answer
Indeed, lost 20 minuts on that, thank you very much :) – VsMaX Aug 26 '13 at 18:42
your welcome... – draks ... Aug 26 '13 at 18:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.