I'm attempting to solve the following question:
What is the volume of the region formed when the ellipse $9x^2+4y^2=36$ is revolved around the line $2x+y=5$?
My try: $$9x^2+4y^2=36$$ $$y =\pm\frac{3}{2}\sqrt{4-x^2}$$ Then I proceeded to draw rectangles from the ellipse onto the line, which had a length of (using the point-to-line distance formula): $$\frac{|2x\pm\frac{3}{2}\sqrt{4-x^2} - 5|}{\sqrt{5}}$$ Using the disk method I obtain as the volume $$\pi\int_{-2}^{2}\left(\frac{|2x\pm\frac{3}{2}\sqrt{4-x^2} - 5|}{\sqrt{5}}\right)^2dx$$ Is this correct? If not, how would I solve this question?
As achille hui points out above, Pappus's Theorem gives the easiest way to solve this problem.
This gives $\displaystyle V=(2\pi\rho)(A)=2\pi\left(\frac{5}{\sqrt{5}}\right)(\pi\cdot2\cdot3)=12\pi^{2}\sqrt{5}.$
The following solution, using the disc method, is definitely harder:
The perpendicular line $y=\frac{1}{2}x+t$ intersects the ellipse where $9x^2+4(\frac{1}{2}x+t)^2=36$,
so expanding gives $10x^2+4xt+4t^2=36\;\;$ or $\;\;5x^2+2tx+2(t^2-9)=0$.
Then $\displaystyle x=\frac{-t\pm3\sqrt{10-t^2}}{5}$, so using $y=\frac{1}{2}x+t$ gives the points of intersection
$\displaystyle\left(\frac{-t-3\sqrt{10-t^2}}{5}, \frac{9t-3\sqrt{10-t^2}}{10}\right), \left(\frac{-t+3\sqrt{10-t^2}}{5}, \frac{9t+3\sqrt{10-t^2}}{10}\right)$ with the ellipse.
Using the formula $\displaystyle d=\frac{\lvert 2x_1+y_1-5\rvert}{\sqrt{5}}$ for the distance from a point $(x_1,y_1)$ to the line $2x+y=5$
and the fact that $\Delta u=\frac{2}{\sqrt{5}}\Delta t$ by similar triangles, where $u$ measures distance along the line $2x+y=5$,
$\displaystyle V=\int_{-\sqrt{10}}^{\sqrt{10}}\pi\left[\left(\frac{10-t+3\sqrt{10-t^2}}{2\sqrt{5}}\right)^2-\left(\frac{10-t-3\sqrt{10-t^2}}{2\sqrt{5}}\right)^2\right]\cdot\frac{2}{\sqrt{5}}dt$
$\;\;=\displaystyle\frac{\pi}{20}\int_{-\sqrt{10}}^{\sqrt{10}}(20-2t)\cdot6\sqrt{10-t^2}\cdot\frac{2}{\sqrt{5}}dt$
$\;\;=\displaystyle\frac{6\pi}{5\sqrt{5}}\int_{-\sqrt{10}}^{\sqrt{10}}(10-t)\sqrt{10-t^2}\;dt=\frac{6\pi}{5\sqrt{5}}\left[\int_{-\sqrt{10}}^{\sqrt{10}}10\sqrt{10-t^2}\;dt-\int_{-\sqrt{10}}^{\sqrt{10}}t\sqrt{10-t^2}\;dt\right]$
$\displaystyle\;\;=\frac{6\pi}{5\sqrt{5}}\left[10\cdot\frac{1}{2}\pi(10)-0\right]=\frac{60\pi^{2}}{\sqrt{5}}=12\pi^{2}\sqrt{5}$.
If we use the shell method instead, we have that
The parallel line $y=t-2x$ intersects the ellipse where $9x^2+4(t-2x)^2=36$,
so expanding gives $25x^2-16tx+4(t^2-9)=0$.
Then $\displaystyle x=\frac{8t\pm6\sqrt{25-t^2}}{25}$, so using $y=t-2x$ gives the points of intersection
$\displaystyle\left(\frac{8t+6\sqrt{25-t^2}}{25}, \frac{9t-12\sqrt{25-t^2}}{25}\right), \left(\frac{8t-6\sqrt{25-t^2}}{25}, \frac{9t+12\sqrt{25-t^2}}{25}\right)$ with the ellipse.
Then $h$ is the distance between these points, so $h=\displaystyle\frac{12}{25}\sqrt{5}\sqrt{25-t^2}$;
and $r$ is the distance from the line $y=t-2x$ to the line $2x+y=5$, so $\displaystyle r=\frac{|2(0)+t-5|}{\sqrt{5}}=\frac{5-t}{\sqrt{5}}$.
Since $\Delta v=\frac{1}{\sqrt{5}}\Delta t$ by similar triangles, where $v$ measures distance from the line $2x+y=5$,
$\displaystyle V=\int_{-5}^{5}2\pi\left(\frac{5-t}{\sqrt{5}}\right)\left(\frac{12}{25}\sqrt{5}\sqrt{25-t^2}\right)\cdot\frac{1}{\sqrt{5}}dt$
$\;\;=\displaystyle\frac{24\pi}{25\sqrt{5}}\int_{-5}^{5}(5-t)\sqrt{25-t^2}\;dt=\frac{24\pi}{25\sqrt{5}}\left[\int_{-5}^{5}5\sqrt{25-t^2}\;dt-\int_{-5}^{5}t\sqrt{25-t^2}\;dt\right]$
$\displaystyle\;\;=\frac{24\pi}{25\sqrt{5}}\left[5\cdot\frac{1}{2}\pi(25)-0\right]=12\pi^{2}\sqrt{5}$.