How to Calculate the Volume of Revolution without the Shell method or the Disk method 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. 
 A: 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 $4$. 
A: A few other possibilities (since you're looking for a challenge):


*

*When you use the washer or shell method, the expressions that create the washers or shells are integrals themselves.  This would make the calculation a triple integral.

*Do this in cylindrical coordinates with the long axis being the $x$ axis.

*This one I don't know for sure, but there are other orthogonal coordinate systems.  One such system might have one set of parameters that matches your outer surface exactly.  If one such exists, find that one and calculate the volume in that coordinate system.  You'll need to calculate the volume element with the Jacobian and other stuff.

