My friend has asked me to look over a physics homework problem.
I've attempted this multiple times, and I have absolutely no idea what I'm doing wrong. My idea was that I have to use the parallel axis theorem in order to relate two integrals and solve for the value of B.
I know the integral from the left end is the following:
$$ \int _0 ^L {x^2 B (L^2 - x^2)} dx = \frac {2BL^5} {15} $$
And I believe the integral from the middle would be the following:
$$ 2 \int _{L/2} ^L {(\frac L 2 - x)^2 B (L^2 - x^2)} dx = \frac {3BL^5} {160} $$
So the parallel axis theorem should be able to relate these two in the following way:
$$ \frac {2BL^5} {15} = \frac {3BL^5} {160} + M(\frac L 2)^2 $$
Which gives a B value of $\frac {24M} {11L^3}$ which can be substituted into the left integral:
$$ \frac {2(\frac {24M} {11L^3})L^5} {15} = \frac {16ML^2} {55} $$
Unfortunately, this is not an answer choice to the problem, and I've tried it many times. I have no idea what I'm doing wrong and any pointers would be much appreciated, thank y'all.
You do not need to do the integral from the middle or use the parallel axis theorem. Your first integral is the answer, since $x$ is the distance to the $Y$-axis and you are correctly computing $\int x^2dM$. All you need is to relate $B$ to $M$. You know that the total mass is $M$, that is $$ \int_0^LB(L^2-x^2)\,dx=M, $$ which gives $$ B=\frac{3M}{2L^3}. $$ Plugging in this value into your first formula, you get $$ I=\frac{ML^2}{5}. $$