Find the volume $V$ of the solid obtained by rotating the region bounded by the given curves about the specified line.
$y = \ln{7x}$;
$y = 3$
$y = 4$
$x = 0$
Rotated about the $Y$-Axis.
How would I approach solving this problem. I'm stuck. Thanks for the help!
Best Answer
Since we need to rotate about the $Y$-axis, we are forming infinitesimal disks which are aligned on the $y$-axis. (Imagine that the $y$-axis is a rod that goes through each center of a thin disk). These disks go from $y=3$ to $y=4$. The $x=0$ is just the $y$-axis.
Geometrically, these thin disks have a radius perpendicular to the $y$-axis and so to find this radius, we can solve $y=\ln{7x}$ as a function of $x$. By doing this, we can use the distance from the $y$-axis to the curve as the radius. (Go ahead and turn your paper counter-clockwise to show the "heights" from the $y$-axis to the function. These "heights" are the radii of your disks.
We solve for $x$ and get: $$y=\ln{7x}$$ $$e^y=e^{\ln{7x}}$$ $$e^y=7x$$ $$x=\frac{1}{7}e^y$$
We are summing up the volume of all these REALLY thin disks. So we are adding each infinitesimal disk between $3$ and $4$ on the $y$-axis. This is the same as integrating a volume function from $3$ to $4$ of an area function at each point. Since rotating this gives us a circle, $\pi{r^2}$.
Our function $x=\frac17e^y$ is now a function that outputs a radius. (Quite convenient but not a coincidence!)
So we integrate and get $$V = \int_3^4{\pi(\frac17e^y)^2}dy$$ $$V = \frac{\pi}{49}\int_3^4e^{2y}dy$$ $$V= \frac{\pi}{49}\left[\frac12e^{2y}\right]_3^4$$ $$V= \frac{\pi}{98}\left(e^8-e^6\right)\approx82.628$$