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

#### MarkFL

Staff member
Here is the question:

What is the surface area of this torus?

Here are the dimensions:

R=13"
r=4"

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.

#### MarkFL

Staff member
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:

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

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

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:

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