# Thread: Finding Volume of a Torus by revolving a disk.

1. ## Finding Volume of a Torus by revolving a disk.

Ok everyone, new to the forum here. I just started taking Calc 2 (online) after graduating 8 years ago. Taking it to apply for master's in physics, so I really need a good grasp here. So I'm a little rusty and it's hard to get help in an online class. Anyways, I've been doing well with understanding how to find volumes by revolution but then I got to this problem and I'm stumped. Is there anyone that can give me a detailed, step by step instruction on how to go about it? I found the answer online, but I'm not just trying to get a good grade, I really need to understand this. That would be awesome. Anyways, here's the problem.

(also, I know there are other ways to do it, but I've been asked to use the washer method, so I'd like to use that).

The disk x2+y2a2 is revolved about the line x=b (b>a) to generate a torus. Find its volume.

2. ## Re: Finding Volume of a Torus by revolving a disk.

The washer method is just a special case of taking slices.
Slice the torus perpendicular to the x-y plane, going from y=-a to y=a.
Each slice is an annulus = a disk with an inner disk removed.

At height $y$, the outer radius is the distance from $(b, y)$ to $(x_1, y)$, where $(x_1, y)$ is on the original circle with $x_1 <0$.
At height $y$, the inner radius is the distance from $(b, y)$ to $(x_2, y)$, where $(x_2, y)$ is on the original circle with $x_2 >0$.

By symmetry, it's obvious that $x_2 = -x_1$.
Thus Outer Radius = $b+x_2$. Inner Radius = $b-x_2$.

$Area(y) = \pi ( (b+x_2)^2 - (b-x_2)^2 )$

$= \pi ( (b^2 + 2bx_2 + x_2^2) - (b^2 - 2bx_2 + x_2^2))$

$= \pi ( 4bx_2 ) = 4 \pi b x_2$.

But $(x_2, y)$ on the circle, $x_2 >0$, so $x_2 = \sqrt{a^2 - y^2}$.

Thus $Area(y) = \pi ( 4bx_2 ) = 4 \pi b \sqrt{a^2 - y^2}$.

Thus $Volume = \int_{-a}^{a} Area(y) dy = 4 \pi b \int_{-a}^{a} \sqrt{a^2 - y^2} dy$.

Now $y = a\sin(t). \ dy = a\cos(t)dt, \ y=-a$ when $t = -\frac{\pi}{2}, \ y = a$ when $t = \frac{\pi}{2}$.

$Volume = 4 \pi b \int_{t = -\frac{\pi}{2}}^{t = \frac{\pi}{2}} \sqrt{a^2 - (a\sin(t))^2} (a\cos(t)dt)$

$= 4 \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 - \sin^2(t)} \cos(t) dt$

$= 4 \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\cos(t)| \cos(t) dt$ (Note $\cos(t) \ge 0$ on this interval.)

$= 4 \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(t) dt$

$= 4 \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos(2t)}{2} dt$

$= 2 \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\cos(2t)) dt$

$= 2 \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dt + 2 \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2t) dt$

$= 2 \pi a^2 b \left\ t \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} + \pi a^2 b \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2t) d(2t)$

$= 2 \pi a^2 b \left( \left(\frac{\pi}{2}\right) - \left(- \frac{\pi}{2}\right) \right) + \pi a^2 b \left\ \sin(2t) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$= 2 \pi a^2 b (\pi) + \pi a^2 b ( \sin(2(\pi/2)) - \sin(2(-\pi/2))$

$= 2 \pi^2 a^2 b + \pi a^2 b ( 0 - 0 )$

$= 2 \pi^2 a^2 b$

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### the volume of x^2 y^2 a^2; about x=b(b>a)

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