Double integral and wrong answer

**totalnewbie** · June 1st 2008, 09:40 AM

As you see from the picture I got 8*pi. Book gives me 4*pi.
Is my answer wrong or right?

**flyingsquirrel** · June 1st 2008, 10:19 AM

Hi

$\int_0^{2\pi}\mathrm{d}\varphi\int_0^2(1-2\rho \cos \varphi-3\rho \sin \varphi)\rho\,\mathrm{d}\rho=\int_0^{2\pi} \left[\frac{\rho^2}{\color{red}2}-2\frac{\rho^3}{3} \cos \varphi-3\frac{\rho^3}{3} \sin \varphi \right]_0^2\mathrm{d}\varphi$

and the result is $4\pi$ .

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