Findinf volume enclosed using triple integrals I have to find the volume cut off from the paraboloid $4z=x^2+y^2$ by the plane $z=4$
I understand that the paraboloid will have its vertex at the orgin and its axis as the $z-axis$. now, if it is cut off by the plane $z=4$, i get the following limits for the triple intgral
$z$ from $0$ to $4$
$x$ from $-\sqrt{16-y^2}$ to $\sqrt{16-y^2}$
$y$ from $-16$ to $16$
However, this is double the correct answer. Where am i going wrong ?

We can interpret this volume as $$\int_0^4 \text{area of disk at height } z\ dz.$$  More formally, we convert to a cylindrical coordinate system, $$\int_0^4 \underbrace{\int_0^{2\pi} \int_0^{\sqrt{4z}} r\ dr\ d\theta}_{\text{area of disk}}\ dz.$$
The area of a disk is $\pi r^2$ where $r$ is the radius, in this case $r=\sqrt{4z}$, so the area is $4\pi z$, and the integral becomes $$\int_0^4 4\pi z\ dz=\left[2\pi z\right]_0^4=32\pi.$$

Alternatively, we can find the volume of solid of revolution via $$2\pi \int_0^{4} x|f(x)-g(x)|\ dx$$ where $f(x)=4$ and $g(x)=\tfrac{1}{4}x^2$.  This gives the volume
\begin{align*}
2\pi \int_0^{4} 4x-\tfrac{1}{4}x^3\ dx &= 2\pi \left[2x^2-\tfrac{1}{16} x^4\right]_0^{4} \\
 &= 2\pi(2 \times 4^2-\tfrac{4^4}{16}) \\
 &= 32\pi.
\end{align*}
