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roisin's question at Yahoo! Answers regarding the volume of a torus

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MarkFL

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Feb 24, 2012
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Here is the question:

Volume by integration help please?

A circle of radius a is centered on the x–axis at the point with co–ordinates
(b, 0), where b > a > 0. The circle is rotated around the y–axis. Determine
(with proof) the volume of the solid so generated.
I have posted a link there to this thread so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello roisin,

The equation describing the circle to be rotated is:

\(\displaystyle (x-b)^2+y^2=a^2\)

Now, so that we know what our goal is, using the formula for the volume of a torus, we should expect to find the volume of the solid of revolution to be:

\(\displaystyle V=2\pi^2a^2b\)

Let's use both the washer and shell methods. We'll begin with the washer method.

Washer Method

The volume of an arbitrary washer is:

\(\displaystyle dV=\pi\left(R^2-r^2 \right)\,dy\)

where:

\(\displaystyle R=b+\sqrt{a^2-y^2}\)

\(\displaystyle r=b-\sqrt{a^2-y^2}\)

Hence:

\(\displaystyle R^2-r^2=(R+r)(R-r)=(2b)\left(2\sqrt{a^2-y^2} \right)=4b\sqrt{a^2-y^2}\)

And so the volume of the arbitrary washer can be written as:

\(\displaystyle dV=4\pi b\sqrt{a^2-y^2}\,dy\)

Now, summing the washers, we get the volume of the solid:

\(\displaystyle V=4\pi b\int_{-a}^a \sqrt{a^2-y^2}\,dy\)

Using the even-function rule, this becomes:

\(\displaystyle V=8\pi b\int_0^a \sqrt{a^2-y^2}\,dy\)

Using the substitution:

\(\displaystyle y=a\sin(\theta)\,\therefore\,dy=a\cos(\theta)\,d \theta\)

We have:

\(\displaystyle V=8\pi b\int_{\theta(0)}^{\theta(a)}\sqrt{a^2-a^2\sin^2(\theta)}\,a\cos(\theta)\,d\theta\)

Now, to change the limits of integration, observe we have:

\(\displaystyle \theta(x)=\sin^{-1}\left(\frac{y}{a} \right)\) and so:

\(\displaystyle \theta(0)=\sin^{-1}\left(\frac{0}{a} \right)=0\)

\(\displaystyle \theta(a)=\sin^{-1}\left(\frac{a}{a} \right)=\frac{\pi}{2}\)

Now, on the interval \(\displaystyle \left(0,\frac{\pi}{2} \right)\), we have the sine and cosine functions being non-negative, hence we may write the integral as:

\(\displaystyle V=8\pi b\int_{0}^{\frac{\pi}{2}}a\sqrt{1-\sin^2(\theta)}\,a\cos(\theta)\,d\theta\)

\(\displaystyle V=8\pi a^2b\int_{0}^{\frac{\pi}{2}}cos^2(\theta)\,d\theta\)

Now, using the identity \(\displaystyle \cos^2(x)=\frac{1+\cos(2x)}{2}\) we have:

\(\displaystyle V=4\pi a^2b\int_{0}^{\frac{\pi}{2}}1+\cos(2\theta)\,d \theta\)

Hence:

\(\displaystyle V=4\pi a^2b\left[\theta+\frac{1}{2}\sin(2\theta) \right]_{0}^{\frac{\pi}{2}}=4\pi a^2b\left(\frac{\pi}{2} \right)=2\pi^2a^2b\)

Okay, this is the result we expected. Now let's look at the shell method.

Shell Method

The volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x\)

\(\displaystyle h=2\sqrt{a^2-(x-b)^2}\)

And thus the volume of the arbitrary shell is:

\(\displaystyle dV=4\pi x\sqrt{a^2-(x-b)^2}\,dx\)

Summing up the shells, we get the volume of the solid:

\(\displaystyle V=4\pi\int_{b-a}^{b+a} x\sqrt{a^2-(x-b)^2}\,dx\)

Using the substitution:

\(\displaystyle x-b=a\sin(\theta)\,\therefore\,dx=a\cos(\theta)\, d\theta\)

we obtain:

\(\displaystyle V=4\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(a\sin(\theta)+b \right)\sqrt{a^2-a^2\sin^2(\theta)}\,a\cos(\theta)\,d\theta\)

\(\displaystyle V=4\pi a^2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(a\sin(\theta)+b \right)\sqrt{1-\sin^2(\theta)}\,\cos(\theta)\,d\theta\)

\(\displaystyle V=4\pi a^2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(a\sin(\theta)+b \right)\cos^2(\theta)\,d\theta\)

\(\displaystyle V=4\pi a^3\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(\theta)\cos^2(\theta)\,dx+4\pi a^2b\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(\theta)\, d\theta\)

Using the odd and even function rules, we obtain:

\(\displaystyle V=8\pi a^2b\int_{0}^{\frac{\pi}{2}}\cos^2(\theta)\, d\theta\)

Now, using the identity \(\displaystyle \cos^2(x)=\frac{1+\cos(2x)}{2}\) we have:

\(\displaystyle V=4\pi a^2b\int_{0}^{\frac{\pi}{2}}1+\cos(2\theta)\,d \theta\)

Hence:

\(\displaystyle V=4\pi a^2b\left[\theta+\frac{1}{2}\sin(2\theta) \right]_{0}^{\frac{\pi}{2}}=4\pi a^2b\left(\frac{\pi}{2} \right)=2\pi^2a^2b\)

Here we have also obtained the desired result.