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Math Help - Finding Volume of a Torus by revolving a disk.

  1. #1
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    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.

    Thanks in advance.
    Last edited by yomikeya; October 4th 2012 at 12:55 PM.
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  2. #2
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    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 (Outer Radius)^2 - pi (Inner Radius)^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
    Last edited by johnsomeone; October 5th 2012 at 11:30 AM.
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