Finding line that divides rotational volume in half I need to rotate the area between the curve $y=x^2$ and $y=9$, bounded in the first quadrant, around the vertical line $x=3$. I then must find the height (m) of the horizontal line that divides the resulting volume in half.
I've been trying to set up two integrals with washers. One from $\int_{m}^{9}3^2-(3-x)^2dy$ and the other for the bottom region $\int_{0}^{m}3^2-(3-x)^2dy$. The $3$ is the outer radius of the washer and the $3-x$ gives the inner hole of the washer. I can't seem to get the right answer.
$\pi\int_{m}^{9}(6\sqrt{y}-y)dy = \pi (4y^\frac{3}{2} - \frac{1}{2}y^2)$ evaluated from 9 to m $= \pi (\frac{135}{2} - (4m^\frac{3}{2} -\frac{1}{2}m^2)$. Similarly, for the bottom region integral, I get $\pi (4m^\frac{3}{2} -\frac{1}{2}m^2)$. I then try to set the volumes equal to each other and solve for m.
I believe the answer I should get is $\frac{9}{\sqrt[3]{2}}$ but I do not get that value for m.

As you've determined, you are wanting to solve the equation $$\pi\left(\frac{135}2-\left(4m^\frac32-\frac12m^2\right)\right)=\pi\left(4m^\frac32-\frac12m^2\right),$$ or equivalently, $$4m^\frac32-\frac12m^2=\frac{135}4.\tag{$\star$}$$
Observe that the function $m\mapsto 4m^\frac32−\frac12m^2$


*

*is defined and continuous on the interval $[0,36],$

*is strictly increasing on the interval $(0,36)$ (since its derivative is positive there),

*takes on a value of $0$ (less than $\frac{135}4$) when $m=0,$

*takes on a value of $216$ (greater than $\frac{135}4$) when $m=36.$


Hence, by Intermediate Value Theorem, there is a unique solution $m$ to
$(\star)$ in the interval $(0,9)$ in particular. At that point, we're done. Finding a nice expression for this $m$ will be prohibitively difficult without a calculator, since it involves finding a solution to a quartic equation. Wolfram|Alpha gives the following form: $$m = 16+\frac12\sqrt\alpha-\frac12\sqrt{2802-\alpha+\frac{56896}{\alpha}}$$
where $$\alpha=934+\frac1{12}\sqrt[3]{14864601600-6046617600\sqrt{6}}+18\cdot 5^\frac23\sqrt[3]{59+24\sqrt{6}}.$$ It's a rather unfriendly-looking number, and unless you know methods for finding roots of a quartic equation, you're not likely to find it without a calculator.
