The region bounded by the parabola $y=L·x·(2-x)$ and by the lines $y=0$ and $x=1$ is shown below.
https://i.stack.imgur.com/BP0L2.jpg
When this region is revolved about the line $y=0$ a solid of revolution is obtained, and when this region is revolved about the line $x=1$ another solid of revolution is obtained. For which value(s) of $L≥0$ do these two solids have the same volume?
a) $L=0$
b) $L=0,L=15/16$
c) $L=0,L=15/14$
d) $L=0,L=13/12$
e) None of the above.
I'm not sure how to approach this problem. I tried solving the integrals for each direction and setting them equal to each other but that doesn't seem to work.
I suggest that you use Pappus's ($2^{nd}$) Centroid Theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2\pi R$. The bottom line is that the volume is given simply by $V=2\pi RA$.
Now, we can show from the definitions of area and centroid that the area is proportional to $L$ and the centroid is independent of $L$ in one direction and not the other. That is,
$$A=\int f(x)\ dx \sim L\\ R_x=\frac{1}{A}\int x\ f(x)\ dx \sim \frac{L}{L}\\ R_y=\frac{1}{2A}\int f^2(x)\ dx \sim L$$
Then, the volume about $x=1$ is $V_{x=1}=2\pi (1-R_x)A$ and the volume about $y=0$ is $V_{y=0}=2\pi R_y A$.
We can calculate from the above equations that $R_x=5/8$ and $R_y=4/10\cdot L$ and then equate $1-R_x=R_y$ to determine $L=15/16$.