How would I find the equation of f(x) in terms of x and y?

Solution 1:

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.

$\int_{-4}^4 \int_{-\sqrt{16-x^2}}^\sqrt{16 - x^2}\int_{-\sqrt{f^2(x) - y^2}}^{\sqrt {f^2(x) - y^2}} dz\ dy\ dx$

Solution 2:

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 $$ \int_{-4}^4\int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}} z(x,y)dy dx $$