A triple integral problem I am having difficulty trying to solve this triple integral problem:
$$\iiint_V \sqrt {x^2+y^2} \sin(z^2)\,dx\,dy\,dz$$
where 
$$V= \left\lbrace (x,y,z)\,\Big|\,\sqrt {x^2+y^2} \le z \le 3\right\rbrace.$$
I am thinking of making $z\,dz$ appear in the integral but cannot figure out how to do it. Could anyone help me?
Thanks in advance!
 A: You are integrating over a solid cone. For any $z_0\in[0,3]$, the section $z=z_0$ is a circle with radius $z_0$ and area $\pi z_0^2$, and the integral of $\sqrt{x^2+y^2}$ over such circle is just
$$ \int_{0}^{z_0} \rho\cdot 2\pi \rho\,d\rho = \frac{2\pi}{3}z_0^3.$$
In particular the original integral equals
$$ \frac{2\pi}{3}\int_{0}^{3} z^3 \sin(z^2)\,dz \stackrel{z\mapsto\sqrt{t}}{=}\frac{\pi}{3}\int_{0}^{9}t\,\sin(t)\,dt =\color{red}{\frac{\pi}{3}(\sin 9-9\cos 9)}.$$
A: If you know how to use polar coordinates in integration, it would be a simple problem.
Observe that the region $\sqrt{x^2 + y^2} \leq z$ corresponds a circle whose original point is $(0,0)$ and radius is $z$. To demonstrate it more clearly, I'll use $f$ to denote the integrand, 
$$ \iint_{\sqrt{x^2 + y^2} \leq z} f(x,y)dxdy = \int_0^{2\pi} d\theta \int_0^z rf(r\cos(\theta),r\cos\theta)dr $$
The above equation holds. To verify, take a transformation $x = r\cos\theta, y = r\sin\theta$. Then $r \in [0,z], \theta \in [0,2\pi]$ (a circle mentioned before), and the Jacobian is just $r$. 
Replacing $f$ with the given integrand, we have 
$$ \iint_{\sqrt{x^2 + y^2} \leq z} \sqrt{x^2 + y^2}dxdy = \int_0^{2\pi} d\theta \int_0^z r^2dr = \int_0^{2\pi}\frac{1}{3}z^3d\theta = \frac{2\pi}{3}z^3 $$
Thefore, plugging it into the original formula, we just get
$$ \frac{2\pi}{3} \int_0^3 \sin(z^2) z^3 dz = \frac{\pi}{3} \int_0^{9} \sin(z)z  dz = \frac{\pi}{3}(\sin{9} - 9 \cos{9}) $$
