I need general help in solving for the area bounded by $y^2+z^2=x$ and $x=y,\ z=0$.
I'm trying to get the limits of integration for $\int \int \int dzdxdy $. Here's my attempt so far:
$0\leq z\leq \sqrt {x-y^2}$
$ \sqrt {x} \leq y\leq x$
$ 0 \leq x\leq 1$
$\int_{1}^{0}\int_{\sqrt{x}}^{x} \int_{0}^{\sqrt{x-y^2}} dzdydx=S$
But I can't solve this integral. Any suggestions?
\begin{align*} V&=\int_{x=0}^{x=1}\int_{y=x}^{y=\sqrt{x}}\int_{z=0}^{z=\sqrt{x-y^2}}\mathrm dx\,\mathrm d y\,\mathrm d x\\ &=\int_{x=0}^{x=1}\int_{y=x}^{y=\sqrt{x}}\sqrt{x-y^2}\,\mathrm d y\,\mathrm d x \end{align*} Now, in order to integrate this we can use a $\color{blue}{\text{trigonometric sustitution}}$ by setting $y=\sqrt{x}\sin t$, $\mathrm d y=\sqrt{x}\cos t\,\mathrm dt$, then \begin{align*} \int\sqrt{x-y^2}\,\mathrm d y&=\int \sqrt{x-x\sin^2 t}\sqrt{x}\cos t\,\mathrm dt\\ &=\int x\cos^2 t\,\mathrm d t\\ &=x\left(\frac{t}{2}+\frac{1}{4}\sin 2t\right)+C\\ &=x\left(\frac{t}{2}+\frac{1}{2}\sin t\cos t\right)+C\\ &=\frac{x}{2}\left(\sin^{-1}\frac{y}{\sqrt{x}}+\frac{y\sqrt{x-y^2}}{x}\right)+C\\ \int_{y=x}^{y=\sqrt{x}}\sqrt{x-y^2}\,\mathrm d y&=\frac{\pi}{4}x-\frac{1}{2}x\sin^{-1}\sqrt{x}-\frac{1}{2}x\sqrt{x-x^2} \end{align*}
It follows \begin{align*} V&=\int_0^1\left(\frac{\pi}{4}x-\frac{1}{2}x\sin^{-1}\sqrt{x}-\frac{1}{2}x\sqrt{x-x^2}\right)\,\mathrm d x \end{align*}