Find solid's volume with double integral?

Question

Calculate the volume of solid B which is bounded by:

1. $x^2+y^2=16$
2. $y+z=6$
3. $z=0$

And $B$ is located in the first octant ($x,y,z>0$).

Answer 1

Use cylindrical coordinates $(\rho,\theta,z)$ with $x=\rho\cos\theta$ and $y=\rho\sin\theta$. Then $$\begin{align*} &x^2+y^2=16\Rightarrow \rho=4 \quad \mbox{(cylinder with radius 4 and $z$ axis)},\\ &y+z=16\Rightarrow z=16-\rho\sin\theta \quad \mbox{(plane)},\\ &z=0\Rightarrow z=0 \quad \mbox{(plane)},\\ &x>0,y>0,z>0\Rightarrow 0<\theta< \pi/2,z>0 \quad \mbox{(first octant)}, \end{align*}$$ Hence $$B=\{(\rho,\theta,z)\;:\; 0< \rho\leq 4, 0< \theta< \pi/2, 0< z\leq 6-\rho \sin\theta\}$$ and $$\begin{align*}\mbox{Vol}(B)&=\iiint_B \rho \,d\rho\, d\theta\,dz=\int_{\rho=0}^{4}\int_{\theta=0}^{\pi/2}\int_{z=0}^{6-\rho \sin\theta}\rho\, dz\, d\theta\,\,d\rho\\&=\int_{\rho=0}^{4}\int_{\theta=0}^{\pi/2}(6\rho-\rho^2 \sin\theta)\,d\theta\,d\rho\\&=\int_{\rho=0}^{4}(3\pi\rho-\rho^2)\,d\rho=24\pi-64/3.\end{align*}$$