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I have been trying to find the surface area of the torus generated by the rotation of $(x-R)^2 + y^2 = r^2$ about the y axis. I tried to use the equation: $$\int_a^b2\pi y\sqrt{1+\left(\frac {dy}{dx}\right)^2}dx$$

I know that the derivative of the equation is $$\frac {dy}{dx}= \frac {-(x+R)}{\sqrt{r^2-(x+R)^2}}$$

If I am correct, the limits of integration are $R+r$ and $R-r$. When I plug it in the equation I get:

$$2\pi\int_{R-r}^{R+r} \sqrt{r^2+(x+R)^2} * \sqrt{1+\frac {(x+R)^2}{r^2-(x+R)^2}}dx $$

When you simplify, the first square root cancels out and in the second square root we are left with r^2: $$2\pi\int_{R-r}^{R+r}rdx$$

Clearly, this isn't going to lead to the right answer of $4\pi^2Rr$. What am I doing wrong?

Bumblebee
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TanMath
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  • It looks like the wrong formula to begin with. – André Nicolas Jun 29 '15 at 01:41
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    Try $2\pi \int _{R-r}^{R+r}xds=2\int _{R-r}^{R+r}x\sqrt{1+\frac {(x+R)^2}{r^2-(x+R)^2}}dx$. This integral gives you the surface area of the upper half only, so you multiply by $2$ when you are done. – Matematleta Jun 29 '15 at 01:42
  • @AndréNicolas why? And you referring to the SA formula? – TanMath Jun 29 '15 at 01:43
  • @Chilango we are only allowed to use this formula... – TanMath Jun 29 '15 at 01:43
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    There are various formulas for surface area. Do you know how they are derived? Your version won't apply because you are integrating with respect to $x$, so the factor of $f(x)$ you have is not right. You are basically calculating a circumference times a small arc length which is $2\pi xds$ and then integrating to find the total SA. – Matematleta Jun 29 '15 at 01:45
  • A little bit of length goes around a circle of radius $x$, not $y$. – André Nicolas Jun 29 '15 at 01:46
  • @AndréNicolas can you please be more clear – TanMath Jun 29 '15 at 01:46
  • @Chilango our prof. hasn't gone over the derivation yet, but he just gave us this formula.. – TanMath Jun 29 '15 at 01:47
  • Probably for rotation about the $x$-axis. – André Nicolas Jun 29 '15 at 01:48
  • @AndréNicolas Oh, that's what you mean? would the surface area be the same if it is rotated around the x-axis or y-axis? I assumed that the y-axis is what I am supposed to rotate the circle around – TanMath Jun 29 '15 at 01:49
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    The surface areas would be quite different. The formula you quoted is the formula for rotating about the $x$-axis. You can use it, interchanging the roles of $x$ and $y$. Or you use the idea behind the formula, as suggested by Chilango. I think interchanging may be easiest. – André Nicolas Jun 29 '15 at 02:38

1 Answers1

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The formula you quoted is for rotating about the $x$-axis. To modify it for rotation about the $y$-axis, interchange the roles of $x$ and $y$. So the surface area, because of symmetry, is $$2\int_{y=0}^r 2\pi x\sqrt{1+\left(\frac{dx}{dy}\right)^2} \,dy.$$ We have $x-R=\pm\sqrt{r^2-y^2}$ (the "inner" part of the circle is $x=R-\sqrt{r^2-y^2}$ and the "outer" part of the circle is $x=R+\sqrt{r^2-y^2}$).

Differentiate. We get $\frac{dx}{dy}=\mp\frac{y}{\sqrt{r^2-y^2}}$. Square, add $1$, take the square root. We get after some simplification that the area is $$2\int_{y=0}^r 2\pi\left(R-\sqrt{r^2-y^2}\right)\frac{r}{\sqrt{r^2-y^2}}\,dy+2\int_{y=0}^r 2\pi\left(R+\sqrt{r^2-y^2}\right)\frac{r}{\sqrt{r^2-y^2}}\,dy.$$ (The first integral gives the inner surface area, the second gives the outer surface area.)

The two integrals can be profitably combined, to give $$2\int_{y=0}^r 2\pi\frac{2Rr}{\sqrt{r^2-y^2}}\,dy.$$ Now the integration is straightforward. We get $(2)(2\pi)(2Rr)(\pi/2)$.

Remark: At the beginning, to keep things fully familiar, it might be more comfortable to take the circle $x^2+(y-R)^2=r^2$, and rotate it about the $x$-axis. Then the surface of the torus is the surface generated by the upper half of the circle, that is, $y=R+\sqrt{r^2-x^2}$, plus the area generated by the lower half of the circle. Same calculation, really, but it may seem easier.

André Nicolas
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