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Kuba's question at Yahoo! Answers regarding finding the surface area of a torus

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Staff member
Feb 24, 2012
Here is the question:

What is the surface area of this torus?

Here are the dimensions:


I thought of it as just a cylinder with the length being 26(pi) and the diameter being 8. I just keep on getting different answers every time I try to solve this problem. I got 2052.88 in², but I don't think it is correct.
I have posted a link there to this thread so the OP can view my work.
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  • #2


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Feb 24, 2012
Hello Kuba,

To find the surface area of a torus whose major radius is $R$ and minor radius is $r$ (where $r\le R$), we may revolve the function:

\(\displaystyle f(x)=\sqrt{r^2-(x-R)^2}\)

about the $y$-axis and double the resulting surface of revolution $S$.

Hence, we may state:

\(\displaystyle S=4\pi\int_{R-r}^{R+r}x\sqrt{1+\left[f'(x) \right]^2}\,dx\)

Computing the required derivative, we find:

\(\displaystyle f'(x)=\frac{-(x-R)}{\sqrt{r^2-(x-R)^2}}\)

And so we obtain:

\(\displaystyle S=4\pi\int_{R-r}^{R+r}x\sqrt{1+\left[\frac{x-R}{\sqrt{r^2-(x-R)^2}} \right]^2}\,dx\)

Now, let's employ the substitution:

\(\displaystyle u=x-R\,\therefore\,du=dx\)

and now we may write:

\(\displaystyle S=4\pi\int_{-r}^{r}(u+R)\sqrt{1+\frac{u^2}{r^2-u^2}}\,du\)

Simplifying the radicand, we obtain:

\(\displaystyle S=4\pi r\int_{-r}^{r}\frac{u+R}{\sqrt{r^2-u^2}}\,du=4\pi r\left(\int_{-r}^{r}\frac{u}{\sqrt{r^2-u^2}}\,dx+\int_{-r}^{r}\frac{R}{\sqrt{r^2-u^2}}\,du \right)\)

On the far right the first integrand is odd and the second is even, and so we are left with:

\(\displaystyle S=8\pi rR\int_{0}^{r}\frac{1}{\sqrt{r^2-u^2}}\,du\)

Now, using the substitution:

\(\displaystyle u=r\sin(\theta)\,\therefore\,du=r\cos(\theta)\)

we obtain:

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

Applying the FTOC, we obtain:

\(\displaystyle S=8\pi rR\left(\frac{\pi}{2}-0 \right)\)

\(\displaystyle S=4\pi^2rR\)

This demonstrates that you can indeed treat the torus as a cylinder with respect to its surface area:

\(\displaystyle S=(2\pi r)(2\pi R)=4\pi^2 rR\)

Similarly, see >>>this thread<<< for the derivation using calculus of the volume of a torus, given as:

\(\displaystyle V=2\pi^2r^2R\)

which we may obtain by treating the torus as a cylinder as follows:

\(\displaystyle V=\pi r^2(2\pi R)=2\pi^2r^2R\)

Now, using the given data:

\(\displaystyle r=4\text{ in}\)

\(\displaystyle R=13\text{ in}\)

we find the surface area of the given torus is:

\(\displaystyle S=4\pi^2(4\text{ in})(13\text{ in})=208\pi^2\text{ in}^2\approx2052.87771543\text{ in}^2\)