Given the function $f=e^x$, calculate the volume as it revolves around the $x$-axis from $x=0$ to $x=2$.
Now, I know that it is easy to calculate the volume using the Shell method, but is there another way to do it? Can we 'extrend' it into three dimentions, more functions and use triple and double integrals instead? I want to do it with more integrals (and perhaps more functions) as I find it to be much more satisfying to do so.
Using the disk method the volume is given by
\begin{equation*} V=\pi \int_{0}^{2}\left[ f(x)\right] ^{2}\,dx=\pi \int_{0}^{2}e^{2x}\,dx. \end{equation*}
EDIT in response to OP's comment. Let \begin{eqnarray*} C &=&\left\{ (y,z)\in \mathbb{R}^{2}:0\leq \sqrt{y^{2}+z^{2}}\leq e^{x}\right\} \\ &=&\left\{ (y,z)\in \mathbb{R}^{2}:0\leq y^{2}+z^{2}\leq e^{2x}\right\}. \end{eqnarray*}
Then the volume $V$ is given by the triple integral \begin{equation*} V=\int_{0}^{2}\left(\iint_{C}\,dy\,dz\right)\,dx. \end{equation*}
Using polar coordinates in the $y,x$-plane \begin{equation*} y=r\sin \theta ,z=r\cos \theta \end{equation*}
since the Jacobian of the transformation of coordinates is $J=\left\vert\dfrac{\partial(y,z)}{\partial(r,\theta)}\right\vert=r$ the integral becomes
\begin{eqnarray*} V &=&\int_{0}^{2}\left( \int_{0}^{e^{x}}\int_{0}^{2\pi }r\,d\theta dr\,\right) dx \\ &=&\int_{0}^{2}\left( \int_{0}^{e^{x}}2\pi r\,dr\right) \,dx \\ &=&\int_{0}^{2}\pi e^{2x}\,dx \\ &=& \frac{\pi}{2}(e^4-1), \end{eqnarray*}
as above.
Comment. Since the question deals with a solid of revolution it is natural to use cylindrical coordinates, instead of Cartesian coordinates, otherwise the evaluation of the triple volume is very difficult. Another possibility is to compute the volume of the portion of the solid located in the first octant and multiply the result by $4$.