Use cylindrical coordinates to find the volume of the solid Q. Q is bounded by the graph of $z = 16 - r^2$ and plane $z = 0$

Question

This is a parabola and a plane should I try and convert $z = 16 - r^2$ back to $x$ and $y$ since its already in terms of $R$? I am stuck on this question any help would be appreciated thanks.

Answer 1

You already have your curve in cylindrical coordinates, try graphing the curve so you can see how to set your limits of integration

$z$ will go from $z$ to your function $f(z)=16-r^2$

$r$ will go from $0$ to $4$

$\theta$ will go from $0$ to $2\pi$

So $\int_0^{2\pi} \int_0^4\int_0^{16-r^2}r \quad dz dr d\theta$ multiplying by the jacobian

$$\int_0^{2\pi}\int_0^{4} 16r-r^3 dr d\theta$$ $$\int_0^{2\pi} 64d\theta$$ $$128\pi$$