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a. A torus is formed by revolving the region bounded by the circle $(x-2)^2 + y^2 = 1$ about the y-axis. Use the disk/washer method to calculate the volume of the torus.

Figure given, showing $r=2$ and with centroid at $(2,0)$

b. Use the disk/washer method to find the volume of the general torus if the circle has radius r and its center is R units from the axis of rotation.

For part a, I started by rewriting equation as $x = 2 \pm \sqrt{1-y^2}$. I was using the washer setup, and simplified to $V= 8\pi \int_{-1}^1 \sqrt{1-y^2}dy$.

The answer is given as $4\pi^2$. Just need to figure out the work to get there.

Again, for part b answer is given as $2\pi^2 r^2 R$. I know I need the answer from part a to solve part b. Looking for any help on how to complete the steps that will give me the answer.


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2 Answers 2

I think you reasoned, absolutely correctly, as follows. We have $x=2\pm\sqrt{1-y}^2$. Look at a slice at height $y$, of width "$dy$." The cross-section at height $y$ is a circle of radius $2+\sqrt{1-y^2}$, with a hole of radius $2-\sqrt{1-y^2}$.

The cross sectional area is therefore $$\pi\left(2+\sqrt{1-y^2}\right)^2-\pi\left(2-\sqrt{1-y^2}\right)^2.$$ Simplify, by expanding the two squares. There is some nice cancellation, and we end up with $8\pi\sqrt{1-y^2}$. So the volume of our slice of thickness $dy$ is about $8\pi\sqrt{1-y^2}\,dy$. "Add up" (integrate) from $-1$ to $1$. We conclude that the volume of the torus is $$8\pi\int_{-1}^1 \sqrt{1-y^2}\,dy.$$

It remains to evaluate the integral. There are two ways, one easy and the other harder.

Let's first do it the easy way. Think about $\int_{-1}^1 \sqrt{1-y^2}\,dy$, or, equivalently, about its good friend $\int_{-1}^1\sqrt{1-x^2}\,dx$.

The integral involving $x$ is exactly the integral you would write down if you wanted to find the area of the top half of the circle $x^2+y^2=1$. But the top half of this circle has area $\dfrac{\pi}{2}$. Multiply by $8\pi$, and we are finished!

But for completeness, let's do it the harder way. By symmetry, our integral is $2$ times $\int_0^1\sqrt{1-y^2}\,dy$. Make the substitution $y=\sin t$. Then $$\int_0^1\sqrt{1-y^2}\,dy=\int_0^{\pi/2} \cos^2 t\,dt.$$ One relatively straightforward way to integrate $\cos^2 t$ is to use the double angle identity $\cos 2t =2\cos^2 t-1$ to rewrite $\cos^2 t$ as $\dfrac{\cos t +1}{2}$. Now the integration is not difficult. The definite integral of the $\cos 2t$ part is $0$.

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I have no idea what the washer setup is, but here is a solution for Part (a).

You made a mistake in your $V$ computation.

Draw a picture. Consider a 'thin sliver' of width $dx$ at $x$. If $x \in [1,3]$, then $y$ ranges from $-\sqrt{1-(x-2)^2}$ to $+\sqrt{1-(x-2)^2}$, and rotating this around the $y$ axis will give the incremental volume $dV = 2 \pi x (2 \sqrt{1-(x-2)^2}) dx$. This will give the integral \begin{eqnarray} V &=& 4 \pi \int_1^3 x \sqrt{1-(x-2)^2} dx \\ &=& 4 \pi \int_{-1}^1 (x+2) \sqrt{1-x^2} dx \\ &=& 4 \pi \int_{-1}^1 2 \sqrt{1-x^2} dx \\ & = & 4 \pi 2 \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \cos^2 \theta d\theta \\ &=& 4 \pi^2 \end{eqnarray}

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