Solve Volume of Circle Rotated Around X-axis: Group Take Home

[Pull and Twist]

I am working on a take home with a group of students and we all got different answers for this problem... can anyone look over it and tell me where I am going wrong?? Let me know if any of my work is confusing...

View attachment 3971

 

[MarkFL]

Well, the solid is a torus, and so we know its volume is:

\(\displaystyle V=\pi(2)^2(2\pi(5))=40\pi^2\)

You may wish to look at this thread:

https://mathhelpboards.com/threads/roisins-question-at-yahoo-answers-regarding-the-volume-of-a-torus.7992/

 

[Pull and Twist]

Well, the solid is a torus, and so we know its volume is:

\(\displaystyle V=\pi(2)^2(2\pi(5))=40\pi^2\)

You may wish to look at this thread:

http://mathhelpboards.com/questions-other-sites-52/roisins-question-yahoo-answers-regarding-volume-torus-7992.html?highlight=torus

I looked at that link and I'm confused as to how you incorporate the fact that the circle is centered at (0,5). I'm not really following the setup in that link.

 

[Prove It]

I am working on a take home with a group of students and we all got different answers for this problem... can anyone look over it and tell me where I am going wrong?? Let me know if any of my work is confusing...

The whole circle has equation $\displaystyle \begin{align*} x^2 + \left( y - 5 \right) ^2 = 4 \end{align*}$, so if we use washers, the inner radius is $\displaystyle \begin{align*} y = 5 - \sqrt{4 - x^2} \end{align*}$ and the outer radius is $\displaystyle \begin{align*} y = 5 + \sqrt{4 - x^2} \end{align*}$, so the volume is

$\displaystyle \begin{align*} V &= \pi \int_{-2}^2{ \left( 5 + \sqrt{4 - x^2} \right) ^2 - \left( 5 - \sqrt{4 - x^2 } \right) ^2 \,\mathrm{d}x } \\ &= 2\pi \int_0^2{ \left( 5 + \sqrt{4 - x^2} \right) ^2 - \left( 5 - \sqrt{4 - x^2} \right) ^2 \, \mathrm{d}x } \\ &= 2\pi \int_0^2{ \left[ \left( 5 + \sqrt{4 - x^2} \right) - \left( 5 - \sqrt{4 - x^2} \right) \right] \left[ \left( 5 + \sqrt{ 4 - x^2} \right) + \left( 5 - \sqrt{4 - x^2} \right) \right] \,\mathrm{d}x } \\ &= 2\pi \int_0^2{ 20\,\sqrt{4 - x^2} \,\mathrm{d}x } \\ &= 40\pi \int_0^2{ \sqrt{ 4 - x^2} \,\mathrm{d}x } \end{align*}$

Now with the substitution $\displaystyle \begin{align*} x = 2\sin{(t)} \implies \mathrm{d}x = 2\cos{(t)}\,\mathrm{d}t \end{align*}$ and noting that when $\displaystyle \begin{align*} x = 0, t = 0 \end{align*}$ and when $\displaystyle \begin{align*} x = 2, t = \frac{\pi}{2} \end{align*}$, the integral becomes

$\displaystyle \begin{align*} 40\pi \int_0^2{\sqrt{4 - x^2}\,\mathrm{d}x} &= 40\pi \int_0^{\frac{\pi}{2}}{ \sqrt{4 - \left[ 2\sin{(t)} \right] ^2 }\,2\cos{(t)}\,\mathrm{d}t} \\ &= 40\pi \int_0^{\frac{\pi}{2}}{\sqrt{4 - 4\sin^2{(t)}}\, 2\cos{(t)}\,\mathrm{d}t} \\ &= 40\pi \int_0^{\frac{\pi}{2}}{ 2\cos{(t)} \, 2\cos{(t)}\,\mathrm{d}t } \\ &= 160\pi \int_0^{\frac{\pi}{2}}{ \cos^2{(t)}\,\mathrm{d}t } \\ &= 160\pi \int_0^{\frac{\pi}{2}}{\frac{1}{2} \left[ 1 + \cos{(2t)} \right] \,\mathrm{d}t} \\ &= 80\pi \int_0^{\frac{\pi}{2}}{1 + \cos{(2t)}\,\mathrm{d}t} \\ &= 80\pi \left[ t + \frac{1}{2}\sin{(2t)} \right]_0^{\frac{\pi}{2}} \\ &= 80\pi \left[ \left( \frac{\pi}{2} + 0 \right) - \left( 0 + 0 \right) \right] \\ &= 80\pi \left( \frac{\pi}{2} \right) \\ &= 40 \pi ^2 \end{align*}$

 

[MarkFL]

I looked at that link and I'm confused as to how you incorporate the fact that the circle is centered at (0,5). I'm not really following the setup in that link.

That problem is in more general terms, and is rotated about the $y$-axis instead, but the method would be essentially the same. :D

 

[Pull and Twist]

Thank you guys. That helped me immensely.

After thinking about it I realized where I was going wrong. I was thinking that I could bisect the torus at y=5 and solve for half the volume and then double it.

I finally realized that the volume of the inner part would actually be less then that of the outer part as it travels less distance to complete the full rotation. I had to imagine cutting a doughnut to visualize it.