Finding Volume of Torus by Revolving Semi-Circle

[icesalmon]

Homework Statement

A torus is formed by revolving the region bounded by the circle x² + y² = 1 about the line x = 2. Find the volume of the solid.

The Attempt at a Solution

I've actually got the answer correct. I'm using shell method. My height, h(x), I believe is double the area of the semi-circle, I have 2(1-x²)^1/2 and my radius p(x) was 1. I'm integrating over the interval [-1,1] and I got 4pi² I'm curious if this was the right thing to do or if it makes sense to more knowledgeable people out there. Thanks

 

[Mark44]

Homework Statement

A torus is formed by revolving the region bounded by the circle x² + y² = 1 about the line x = 2. Find the volume of the solid.

The Attempt at a Solution

I've actually got the answer correct. I'm using shell method. My height, h(x), I believe is double the area of the semi-circle, I have 2(1-x²)^1/2 and my radius p(x) was 1. I'm integrating over the interval [-1,1] and I got 4pi² I'm curious if this was the right thing to do or if it makes sense to more knowledgeable people out there. Thanks

If you got the right answer, I don't see how. Using the shell method, a typical volume element would be 2$\pi$ * radius * height * Δx

The height is 2√(1 - x²), as you say, but the radius is not a constant - it is 2 - x. The radius of the circle is 1, but that doesn't have anything directly to do with the volume of the torus, other than playing a role in the height.

 

[icesalmon]

so the height is 2(1-x²)^1/2 and the radius is 2-x
thanks, got it.

 

[Mark44]

Can you show how you did the integration? I'm getting 4$\pi^2$, which is the answer you showed in post 1.

You should break the integral into two integrals. One of them would be somewhat difficult if you actually tried to find an antiderivative, but it's easy when you use the geometry that the integral represents. The other integral is an easy substitution.

 

[icesalmon]

I have 4(1-x²)^1/2 -2x(1-x²)^1/2 for my first integrand after factoring. integrating -2x(1-x²)^1/2 I believe goes to 0. so I'm left with 4(1-x²)^1/2 to integrate so, I factor out the 4 and use trig subs to finish off the integral letting x = sin(θ) my bounds change to [-pi/2,pi/2] and I end up with 8pi/2(theta + sin(theta)) the sin(theta) is zero at both pi and -pi so I have 4pi(pi/2-(-pi/2)) = 4pi(pi) = 4pi²

 

[icesalmon]

i'm not "supposed" to know how to use trigonometric substitution at this point. The (1-x²)^1/2 is the area under the upper portion of the circle but I'm not sure what kind of techniques are being used to reduce it down to pi²

edit: the integrand (1-x²)^1/2 represents half of the area of a circle so I would multiply 8pi by pi(r)²/2 with r = 1
I would get 8pi(pi(1)/2) = 4pi²

 

[Mark44]

i'm not "supposed" to know how to use trigonometric substitution at this point. The (1-x²)^1/2 is the area under the upper portion of the circle but I'm not sure what kind of techniques are being used to reduce it down to pi²

edit: the integrand (1-x²)^1/2 represents half of the area of a circle so I would multiply 8pi by pi(r)²/2 with r = 1
I would get 8pi(pi(1)/2) = 4pi²

Right, that's exactly what you're supposed to do. That's what I meant when I said that the integral was easy if you use the geometry that it represents.

And yes, the other integral's contribution is 0.

 

[icesalmon]

yes, of course. Thank you