Integrable in 3 variables

Question

I have to compute $$\iiint_D(x^2+2y^2)z\, dxdydz$$ with $D=\{x^2+y^2+z^2\leq 4, z>\sqrt{x^2+y^2}\}$.

I tried with some classical change of variables but without more sense. Same ideas?

Answer 1

When you want to integrate over the volume trapped between $z=f(x,y)$ and $z=g(x,y)$ , you integrate over the area bounded by the curve $f(x,y)=g(x,y)$ and the $z$ limit would be from $f(x,y)$ to $g(x,y)$. What I am trying to say is:-

So what you want is :-

$$\iint_{D}\int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{4-x^{2}-y^{2}}}\bigg((x^{2}+2y^{2})z\bigg)\,dz\,dx\,dy$$

where $D$ is the region bounded by the curve $x^{2}+y^{2}=4-x^{2}-y^{2}$ which is just the circular region $x^{2}+y^{2}\leq2$

Now just evaluate:-

$$\iint_{x^{2}+y^{2}\leq 2}(x^{2}+2y^{2})\bigg(2-x^{2}-y^{2}\bigg)\,dxdy$$

Now you use polar coordinates to make life easier.

$$\int_{0}^{\sqrt{2}}\int_{0}^{2\pi}r^{2}\bigg(\cos^{2}(\theta)+2\sin^{2}(\theta)\bigg)(2-r^{2})r\,d\theta\,dr$$ .

$$=(1+\pi)\int_{0}^{\sqrt{2}}r^{2}r(2-r^{2})\,dr$$

Substitute $r^{2}=t$ to have:-

$$\frac{1+\pi}{2}\int_{0}^{2}t(2-t)\,dt = \frac{1+\pi}{2}\cdot\frac{4}{3}=\frac{2(1+\pi)}{3}$$