Question

Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate​ axes, the curve y = e Superscript x​, and the line x = ln 3 about the line x = ln 3.

Answer 1

The bounded region we care about is the set

$\{(x,y)\mid0\le x\le\ln3,0\le y\le e^x\}$

Use the shell method. The attached image shows one shell generated by the revolution. Its height is $e^x$ and its radius is the distance from the labeled point $x$ to the axis of revolution $x=\ln3$, or $\ln3-x$. The area of this shell is then $2\pi(\ln3-x)e^x$. We integrate over $0\le x\le\ln3$ to get the volume:

$\displaystyle2\pi\int_0^{\ln3}(\ln3-x)e^x\,\mathrm dx=2\pi(2-\ln3)\approx5.66$

(The integral can be computed by parts if you don't already know the antiderivative of $xe^x$.)