Let $F:=\left\{(x,y,z) \in \mathbb{R}^{3}\mid 0\leq x\leq 1, y^{2} +z^{2}\leq x^{2}(1-x)^{2} \right\}$. Find the volume.
The bounds that I got are:
- $0\leq x\leq 1$
- $-\sqrt{x^{2}(1-x)^{2}} \leq y\leq \sqrt{x^{2}(1-x)^{2}} \Rightarrow -x+x^{2}\leq y\leq x-x^{2}$
- $-\sqrt{x^{2}(1-x)^{2}-y^{2}} \leq z\leq \sqrt{x^{2}(1-x)^{2}-y^{2}}$
I created a sketch so we have some kind of a circle on the xy-plane and some cone-shaped figure in the positive and negative z regions. I suppose it is enough to calculate the integral on the positive z axis and than to double the result. I tried to integrate using a triple integral, but this integral is too hard. Is there a better way?
The integral in Cartesian coordinates actually isn't so bad.
Integrating with respect to $z$ is trivial:
$$\int_0^1\int_{-x(1-x)}^{x(1-x)}\int_{-\sqrt{x^2(1-x)^2-y^2}}^{\sqrt{x^2(1-x)^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx=2\int_0^1\int_{-x(1-x)}^{x(1-x)}\sqrt{x^2(1-x)^2-y^2}\,\mathrm dy\,\mathrm dx$$
The integral with respect to $y$ is perhaps the trickiest part if you're not familiar with trigonometric substitutions. Take $y=x(1-x)\sin u$ and $\mathrm dy=x(1-x)\cos u\,\mathrm du$. Under this transformation we treat $x$ as fixed, and we get lower and upper limits
$$-x(1-x)=x(1-x)\sin u\implies u=\arcsin(-1)=-\frac\pi2$$
$$x(1-x)=x(1-x)\sin u\implies u=\arcsin1=\frac\pi2$$
so that the integral becomes
$$2\int_0^1\int_{-\pi/2}^{\pi/2}\sqrt{x^2(1-x)^2-\left(x(1-x)\sin u\right)^2}\,x(1-x)\cos u\,\mathrm du\,\mathrm dx$$
$$=2\int_0^1x^2(1-x)^2\left(\int_{-\pi/2}^{\pi/2}\sqrt{1-\sin^2u}\,\cos u\,\mathrm du\right)\,\mathrm dx$$
since $0\le x\le1\implies\sqrt{x^2(1-x)^2}=|x(1-x)|=x(1-x)$.
$\sqrt{1-\sin^2u}=\sqrt{\cos^2u}=|\cos u|=\cos u$ for $-\frac\pi2\le u\le\frac\pi2$, so the inner integral reduces to
$$\int_{-\pi/2}^{\pi/2}\cos^2u\,\mathrm du=\frac12\int_{-\pi/2}^{\pi/2}(1+\cos2u)\,\mathrm du=\frac\pi2$$
The original integral is then equivalent to
$$\pi\int_0^1x^2(1-x)^2\,\mathrm dx=\frac\pi{30}$$
Incidentally, this is precisely the setup for finding the volume of the solid of revolution generated by revolving the region between $y=x(1-x)$ and the $x$-axis about $y=0$.