Volume integration Let
$$D = \left\{(x,y,z)\in\mathbb{R}^{3}\mid x\ge0,0\le y\le x, x^2+y^2\le {16}, 0\le z\le {5}\right\}.$$
I want to integrate
$$\displaystyle\iiint\limits_{D}\left({-4\,z+y^2+x^2}\right)\,\mathrm{d}V $$
We can see that $x^2+y^2=r^2$ so $r^2=16$. $r\to[0,16]$ and $\theta\to[0,2\pi]$ and $z \to[0,5]$
And then integration
$$\int_0^{2\pi}\int_0^5\int_0^{16}(-4z+y^2+x^2)\,dV= \int_0^{2\pi}\int_0^5\int_0^{16}(-4z+r^2)\,drdzd\theta=\frac{36160\pi}{3}$$
That is wrong answer and I don't know where I have done mistake.

There are two errors. 


*

*$r^2=16$ means $r=4$ not $16$

*$x\ge 0$ and $0\le y\le x$ means that $\theta$ varies between $0$ and $\pi/4$
