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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!

image

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+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! –  Michael Cwienczek Aug 26 '13 at 19:10
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1 Answer

up vote 0 down vote accepted

There is something wrong in:

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

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Indeed, lost 20 minuts on that, thank you very much :) –  Michael Cwienczek Aug 26 '13 at 18:42
    
your welcome... –  draks ... Aug 26 '13 at 18:45
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