Good night, i have a serious problem solving this integral.
$\int_{0}^{2}\int_{0}^{\sqrt{2x-x^{2}}}\int_{0}^{a}z\sqrt{x^{2}+y^{2}}dzdydx$
I make a change of cylindrical coordinates, and when i make the change my integral change:
$\int_{0}^{1}\int_{0}^{\varPi}\int_{0}^{a}zr^{2}dzd\theta dr$ but when i computing the integral, the answer is: $\frac{8}{9}a^{2} $ and my integral $\frac{1}{6}a^{2}\varPi$ please help me!
The upper bound for $y$ is $$y=\sqrt{2x-x^2}$$ So $$y^2+x^2-2x=r^2-2r\cos\theta=0$$ So that means the limit on $r$ is $r\le2\cos\theta$. And $0\le\theta\le\frac{\pi}2$ because $y\ge0$ and $x\ge0$. Then we have $$\begin{align}\int_0^2\int_0^{\sqrt{2x-x^2}}\int_0^az\sqrt{x^2+y^2}dz\,dy\,dx&=\int_0^{\frac{\pi}2}\int_0^{2\cos\theta}\int_0^azr\,dz\,dr\,d\theta\\ &=\frac12a^2\cdot\frac13(8)\int_0^{\frac{\pi}2}\cos^3\theta\,d\theta\\ &=\frac43a^2\int_0^{\frac{\pi}2}(1-\sin^2\theta)\cos\theta\,d\theta\\ &=\frac43a^2\left[\sin\theta-\frac13\sin^3\theta\right]_0^{\frac{\pi}2}=\frac89a^2\end{align}$$ EDIT: The situation might be a little more clear if we append a plot of the region of integration.