How would I find the equation of f(x) in terms of x and y? The function $f(x)$ was rotated along the x-axis to form a surface.
$$  \int_{-4}^{4} \int_{y=-\sqrt{16-x^{2}}}^{y=\sqrt{16-x^{2}}} \int_{-f(x)}^{f(x)} dzdydx$$
This is what i thought it would be however, I am not sure what I should be integrating... I have my limits for my integrals however, I do not have the equation of this cylinder to integrate into (which I think I need). If anyone can either fix my integration (if it is wrong) or/and provide me with the equation of a cylinder that would be great.  Or if someone can find my f(x) as f(x,y).
 A: lets put $r = f(x)$
When we rotate the the function around the x axis, the magnitude of the function will be the radius of the rotated body.
$y = r\cos t\\
z = r\sin t\\
t = \arccos \frac {y}{r}\\
z = r\sin(\arccos \frac {y}{r}) = r \sqrt {1 - (\frac yr)^2} = \sqrt{f^2(x) - y^2}$
$dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y}, 1)\\$
$(-\frac {f(x)f'(x)}{\sqrt{f(x)^2 - y^2}}, \frac {y}{\sqrt{f^2(x)-y^2}}, 1)\\
\|dS\| = \sqrt {\frac {f^2(x)f'^2(x) + y^2 + f^2(x) - y^2}{f^2(x) - y^2}}\\
\sqrt {\frac {f^2(x)(f'^2(x)+1)}{f^2(x) - y^2}}$
The surface of inside the cylinder:
$\int_{-4}^4 \int_{-\sqrt{16-x^2}}^\sqrt{16 - x^2} f(x)\sqrt {\frac {f'^2(x) + 1}{f^2(x) + y^2}} \ dy\ dx$
And that would be one of the surfaces... there would also be a surface where $z$ is negative, doubling the result, if you want both.
Update... the work above gives the surface area...
If you want the volume.
$\int_{-4}^4 \int_{-\sqrt{16-x^2}}^\sqrt{16 - x^2}\int_{-\sqrt{f^2(x) - y^2}}^{\sqrt {f^2(x) - y^2}} dz\ dy\ dx$
A: Given your function $f(x)$, the equation of (half) the surface is
$$
z(x,y)=\sqrt{f(x)^2-y^2}
$$
because for any point of the surface the distance from the $x$ axis is $f(x)$.
So the volume over the region $x^2+y^2\le 16$ is
$$
\int_{-4}^4\int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} z(x,y)dy dx
$$
