Find the volume of the figures rotating around the y-axis and bounded by

Question

Find the volume of the figures rotating around the y-axis and bounded by: $$y=e^x+6$$ $$y=e^{2x}$$ $$x=0$$ Here's a chart (hopefully correctly):

What should happen next? What is substitute into this formula? $$V_y=\pi \int _a^b\:y^2dx$$

Answer 1

It is simplest to use the Method of Cylindrical Shells. First let is see where the curves $y=e^{2x}$ and $y=e^x+6$ meet. Let $w=e^x$. We are solving $w^2-w-6=0$. The only positive root is $w=3$, giving $x=\ln 3$.

The "height" of the shell at $x$ is $e^x+6-e^{2x}$, and the radius is $x$. So the volume is $$\int_0^{\ln 3} 2\pi x\left(e^x+6-e^{2x}\right)\,dx.\tag{1}$$ The integration is not too bad, one uses integration by parts.

Remark: We have given a very condensed "explanation" of the components of Expression (1). It may be sufficient if you have some experience in using the Method, but not otherwise.