I have to evaluate the following triple integral:
$$\iiint_{V}(3z^2+y^2)\,dx\,dy\,dz$$
where $V=\{(x,y,z)\in\mathbb{R}^3,\ x^2+y^2\ge z^2,\ 0\le x^2+y^2\le \frac{1}{2}\}$
Passing to the cylindrical coordinates, I find out that $\ 0 \le \rho \le \frac{\sqrt2}{2},\ 0 \le \theta \le 2\pi,\ -\rho \le z \le \rho$ and the new integral will be $$\iiint_{V'}(3z^2+\rho^2\sin^2\theta)\rho\,d\rho\,d\theta\,dz$$
I'm not so sure how to solve this. Is it the product of three simple integrals?
It isn't the product of $3$ simple integrals, as the boundary conditions of $z$ depend on $\rho$.
We can evaluate this as - \begin{align} \iiint_{V'}(3z^2+\rho^2\sin^2\theta)\rho d\rho d\theta dz &= \int_{0}^{\frac{1}{\sqrt 2}}\int_{-\rho}^{\rho}\int_0^{2\pi}(3z^2+\rho^2\sin^2\theta)\rho d\theta dz d\rho \\ &= \int_{0}^{\frac{1}{\sqrt 2}}\int_{-\rho}^{\rho}(6\pi z^2+\pi\rho^2)\rho dz d\rho \\ &= \int_{0}^{\frac{1}{\sqrt 2}}(4\pi \rho^3+2\pi\rho^3)\rho d\rho \\ &= \frac{3\pi}{10\sqrt 2} \end{align}