$D=${$ x^2-4(y^2+z^2)\ge{0} \ ,2x^2+y^2+z^2 \le{1} $}
- So I have a double cone on the x axes
- and an Ellipsoid
I have to compute $\int_{D}|x|(1+yz)dxdydz$
[UPDATE]
I think this is the way to proceed:
$\int_{D}|x|(1+yz)dV = \int_{D}x(1+yz)dV-\int_{D}x(1+yz)dV$
for the first integral, I took the x range :
$x=(range)= [2\sqrt{y^2+z^2},\sqrt{1-y^2/2-z^2/2}]$
In order to simplify the integral I change to cylindrical coordinates:
$x_r =(range)=[2r,\sqrt{1-r^2}/\sqrt{2}]$
eventually : ($1/3$ is the radius of the circle about the intersection of the cone and the ellipsoid)
$\int_{0}^{2\pi}\int_{0}^{1/3}\int_{2r}^{\sqrt{1-r^2}/\sqrt{2}}{x(1+r^2\sin{\theta}\cos{\theta})rdxdrd\theta}$
the integral seems kinda wrong, what do you think ?
Notice that by symmetry of $D$, we have
$$\int_D |x|yz \, dV = 0$$
as the signs of $y$ and $z$ will cancel each other out over different octants.
Hence the quantity of interest is just \begin{align}\int_D |x| \, dV &\end{align}
We can focus on just on the first octant and then multiply it by $8$.
\begin{align} &\int_D |x| \, dV \\&= 8 \int_{0}^{\frac{\pi}2}\int_{0}^{\frac13}\int_{2r}^{\sqrt{\frac{1-r^2}2}}xr\,\,\, dx dr d\theta \\ &=4 \int_{0}^{\frac{\pi}2}\int_0^\frac13r\left(\frac{1-r^2}{2} -4r^2\right)\,\, dr d\theta \\ &=4 \int_{0}^{\frac{\pi}2}\int_0^\frac13r\left(\frac{1-9r^2}{2} \right)\,\, dr d\theta \\ &=\pi \int_0^\frac13r\left(1-9r^2 \right)\, dr \\ &= \pi \left[ \frac{r^2}2-\frac{9r^4}{4}\right]_0^\frac13\\ &= \pi \left[ \frac1{18}-\frac1{36} \right]\\ &= \frac{\pi}{36} \end{align}