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

AI Thread Summary
To find the surface area of a torus with a major radius of 13 inches and a minor radius of 4 inches, the formula used is S = 4π²rR. By substituting the values, the calculation yields a surface area of approximately 2052.88 square inches. The discussion clarifies that treating the torus as a cylinder is valid for this calculation. The derived formula confirms the initial estimate provided by the user. Thus, the surface area of the specified torus is accurately calculated as approximately 2052.88 in².
MarkFL
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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.
 
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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:

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

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

Hence, we may state:

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

Computing the required derivative, we find:

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

And so we obtain:

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

Now, let's employ the substitution:

$$u=x-R\,\therefore\,du=dx$$

and now we may write:

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

Simplifying the radicand, we obtain:

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

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

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

Now, using the substitution:

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

we obtain:

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

Applying the FTOC, we obtain:

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

$$S=4\pi^2rR$$

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

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

Similarly, see http://mathhelpboards.com/questions-other-sites-52/roisins-question-yahoo-answers-regarding-volume-torus-7992.html?highlight=torus for the derivation using calculus of the volume of a torus, given as:

$$V=2\pi^2r^2R$$

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

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

Now, using the given data:

$$r=4\text{ in}$$

$$R=13\text{ in}$$

we find the surface area of the given torus is:

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