Integral using polar co-ordinates

**Showcase_22** · Oct 9th 2009, 10:42 AM

Solve $\int \int_{x^2+y^2 \leq 1} (1+x^2+y^2)^{\frac{1}{3}}~dx~dy$

Substituting $x=r \cos \theta$ and $y=r \sin \theta$ :

$\int_0^{2 \pi} \int_0^1 r(1+r^2)^{\frac{1}{3}}~dr~d \theta$

$\int_0^{2 \pi} \left[ \frac{3}{8}(1+r^2)^{\frac{4}{3}} \right]_0^1~d \theta$

$\int_0^{2 \pi} \frac{3}{8} \left( 2^{\frac{4}{3}}-1 \right)~d \theta$

$=\frac{3 \pi}{4} \left(2^{\frac{4}{3}}-1 \right)$

This doesn't really look like a really good answer. =S

So, is this right?

**Krizalid** · Oct 9th 2009, 11:43 AM

it's correct, so relax.

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