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Good night, i have a problem when i go to verify the integration limit $0\leq\theta\leq\varPi/2$ because i think the integration limit go to $0\leq\theta\leq\varPi$ because is an half a circle.

$\begin{align}\int_0^2\int_0^{\sqrt{2x-x^2}}\int_0^az\sqrt{x^2+y^2}dz\,dy\,dx&=\int_0^{\frac{\pi}2}\int_0^{2\cos\theta}\int_0^azr\,dz\,dr\,d\theta\\ &=\frac12a^2\cdot\frac13(8)\int_0^{\frac{\pi}2}\cos^3\theta\,d\theta\\ &=\frac43a^2\int_0^{\frac{\pi}2}(1-\sin^2\theta)\cos\theta\,d\theta\\ &=\frac43a^2\left[\sin\theta-\frac13\sin^3\theta\right]_0^{\frac{\pi}2}=\frac89a^2\end{align}$

the region of integration is:

Region of integration

please help me with the dude.

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Since your domain of interest is in the first quadrant, it should be from $0$ to $\frac{\pi}{2}$.

The $\theta$ should describe the angle from points of your domain to the origin.

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  • $\begingroup$ thanks, it's very simply! good! $\endgroup$ – Bvss12 May 19 '16 at 1:32

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