So i have a triple integral:
$$\iiint\limits_G (x^2+2y^2){dV}$$
where region G is defined as:
$$(x^2+2y^2+4z^2)^{\frac{3}{2}}=x^2+2y^2$$
So my first idea is to use spherical coordinates since the left part is simmilar to a sphere, but seem a little different.
So i have: $$x=rcos{({\phi})}cos{{(\theta})}$$ $$y=rsin(\phi)cos(\theta)$$ $$z=rsin(\theta)$$
So how can i get bounds of my coordinates?
Normally $phi$ goes from $0$ to $2\pi$ but am not sure how can i check it, i have to visualize the region i guess, but am having problems with this one.
Any help regarding boundaries, calcualting the integral, or correcting my initial ideas is welcome.
Thank you in advance.
Let $x=2r\cos\phi\sin\theta, y=\sqrt{2}r\sin\phi\sin\theta, z=r\cos\theta$. Then from the equation we get: $2r = \sin^2 \theta$.
So our boundaries are $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}; 0 \le \phi \le 2\pi; 0 \le r \le \frac{\sin^2 \theta}{2}$. Now the integral becomes:
$$\iiint\limits_G (x^2+2y^2){dV} = \int_0^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{\frac{\sin^2 \theta}{2}} 8r^3 r^2 \sin \theta \ drd\theta d\phi$$
This should be computable.