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 don't get that value for m.
If I continue, I get $\frac{135}{2}$ = $8m^\frac{3}{2}$ - $m^2$ but I don't see any easy way to solve without using a calculator.
The volume of a rotational solid is $\pi\int r^2\mathrm dx$, where $x$ is an arbitrary variable and $r$ is the radius in terms of $x$. Now, somewhat counterintuitively, let's put $r$ in terms of $y$. For any value of $y\in\Bbb R^+$, $x=\sqrt y$. The radius of the solid is the distance between $x=3$ and $f(y)=\sqrt y$, which is given by $3-\sqrt y$.
Thus, we have $$\begin{align} V&=\pi\int_0^9 (3-\sqrt y)^2\mathrm dy\\ &=\pi\left(\int_0^9(9+y)\mathrm dy-\int_0^96\sqrt y\mathrm dy \right)\\ &=\pi\left(\left.9y+\frac{1}{2}y^2\right|_0^9-\left.4y^{3/2}\right|_0^9\right)\\ &=\pi\left(81+40\frac{1}{2}-108\right)=13.5\pi \end{align}$$
The reason you're off is, as far as I can tell, you're still rotating around $x=0$ instead of rotating about $x=3$, as the question dictates