Self-studying integral calculus and I got this problem:

The circle $x^2+y^2=a^2$ is rotated around a line tangent to the circle. Find the area of the surface of rotation.

I made the construction:

enter image description here

There were a few hints given alongside this question, namely: "Set up coordinate axes and a convenient parametrization of the circle. What does the polar graph $r=2a\sin(\theta)$ look like?" I understood the first and last hints since under this new coordinate system, the circle's equation becomes: $${x_n}^2+(y_n-a)^2=a^2$$ Which when converted to polar, gives you the last hint. However I was unable to describe this in terms of parameters, so I decided to take the upper semicircle's surface area of revolution going from $a$ to $-a$ and multiplying that by 2 to account for the lower semicircle. My integral: $$2\int_{-a}^a 2\pi (\sqrt{a^2-x^2}+a)\sqrt{1+\frac{x^2}{a^2-x^2}} dx$$ Upon simplification: $$4\pi\int_{-a}^a a+\frac{a^2}{\sqrt{a^2-x^2}}dx$$ Evaluation leads me to: $$8\pi a^2 + 4\pi^2 a^2$$

However my book (Serge Lang's First Course in Calculus) gives only $4\pi^2 a^2$. Where has my logic gone wrong if I am getting an extraneous term $8\pi a^2$?

EDIT for clarity on integral setup: I first rearranged for $y$ while taking positive square root as I want to take the upper semicircle into consideration for surface of revolution about x-axis. I'll double this to account for the lower semicircle. This gives: $$y=\sqrt{a^2-x^2}+a$$ Using the surface of revolution formula with the derivative as $$\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}$$ Using this into the surface of revolution integral nets me my first integral in this post (also applied $\times$2)

  • 2
    $\begingroup$ I do not understand what you are doing. Can you please explain your integral setup? If you are rotating about a tangent, you cannot take volume of rotation of a semicircle and multiply by $2$. Volume is a function of the distance from the axis of rotation. $\endgroup$
    – Math Lover
    Jan 25 at 16:15
  • 1
    $\begingroup$ Also the volume of rotation of a circle around any tangent is going to be the same so choose a tangent that is easy to work with $\endgroup$
    – Math Lover
    Jan 25 at 16:16
  • $\begingroup$ That is the formula for surface area of revolution around the x-axis. I'll add the setup but I figured out my issue now using the parameters. $\endgroup$
    – Doobius
    Jan 25 at 16:18
  • $\begingroup$ ok so are you rotating $x^2 + (y-a)^2 = a^2$ around x-axis? $\endgroup$
    – Math Lover
    Jan 25 at 16:20
  • 2
    $\begingroup$ Hint: Theorem of Pappus $\endgroup$
    – robjohn
    Jan 25 at 16:51

3 Answers 3


After changing the coordinates, in effect you are rotating $x^2 + (y-a)^2 = a^2$ around x-axis.

The circle is $x^2 + y^2 = 2 ay$

$ \displaystyle y' = \frac{x}{a-y}$

$ \displaystyle ds = \sqrt{1 + (y')^2} ~dx = \frac{a}{|y-a|} ~ dx$

For lower half -

$y = a - \sqrt{a^2-x^2}$

So, $ \displaystyle S_1 = 2 \pi a \int_{-a}^a \frac{a - \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$
$ = 2 \pi a^2 (\pi - 2)$

For upper half -

$y = a + \sqrt{a^2-x^2}$

So, $ \displaystyle S_2 = 2 \pi a \int_{-a}^a \frac{a + \sqrt{a^2-x^2}}{\sqrt{a^2-x^2}} ~ dx$
$ = 2 \pi a^2 (\pi + 2)$

Adding both, $S = 4 \pi^2 a^2$

But it is easier in polar coordinates as I mentioned in comments. The circle is,

$r = 2a \sin\theta, 0 \leq \theta \leq a$

$\dfrac{dr}{d\theta} = 2a \cos\theta$

$ \displaystyle ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} ~ d\theta = 2a ~ d\theta$

$y = 2a\sin^2\theta$

So the integral is,

$ \displaystyle S = 8 \pi a^2 \int_0^{\pi} \sin^2\theta ~ d\theta = 4 \pi^2 a^2$


In spite of your obfuscating figure, you are asking for the surface area of a torus whose inner radius, $R$ (to the center of the cross-section) and outer radius, $r$ (that of the cross-section) are the same. This is well known to be $S=4\pi^2Rr$ (see, for example the CRC Mathematical Tables). So in your case, $S=4\pi^2a^2$

We can derive this result with Pappus's ($1^{st}$) Centroid Theorem, which states that the surface area $S$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$ and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance $d$ traveled by its geometric centroid. Simply put, $S=2πRL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is the curve length. In your case, $R=a$ and $L$ is the circumference of the circle, i.e., $=2\pi a$, so that $S=4\pi^2a^2.$


Comment on the Question

I believe that this is the standard setup for surface of revolution about the $x$-axis $$ \int_{-a}^a\overbrace{\quad2\pi y\quad\vphantom{\frac{a}{\sqrt{a^2}}}}^{\substack{\text{account for}\\\text{revolution}}}\overbrace{\frac{a}{\sqrt{a^2-x^2}}}^{\mathrm{d}s/\mathrm{d}x}\,\mathrm{d}x $$ However, the part for the upper arc of the circle does not give the same area as that for the lower arc of the circle, so we need to compute both separately: $$ \begin{align} &\int_{-a}^a2\pi\left(a+\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{upper}\\ &+\int_{-a}^a2\pi\left(a-\sqrt{a^2-x^2}\right)\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{lower}\\ &=2\pi\int_{-a}^a2a\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\[6pt] &=4\pi^2a^2 \end{align} $$

Other Approaches

Revolution Around the $\boldsymbol{x}$-axis

As I had posted before I realized the question was revolving around the $y$-axis, if we revolve around the $x$-axis, the upper and lower parts of the surface have the same area, so we can just multiply the upper integral by $2$ in this case. Thus, the formula is $$ \begin{align} 2\int_0^{2a}2\pi x\,\frac{a}{\sqrt{a^2-(x-a)^2}}\,\mathrm{d}x &=2\int_{-a}^a2\pi(x+a)\,\frac{a}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{1a}\\ &=4\pi a^2\int_{-1}^1(x+1)\frac1{\sqrt{1-x^2}}\,\mathrm{d}x\tag{1b}\\ &=4\pi a^2\int_{-\pi/2}^{\pi/2}(\sin(x)+1)\,\mathrm{d}x\tag{1c}\\[9pt] &=4\pi^2a^2\tag{1d} \end{align} $$ Explanation:
$\text{(1a)}$: substitute $x\mapsto x+a$
$\text{(1b)}$: substitute $x\mapsto ax$
$\text{(1c)}$: substitute $x\mapsto\sin(x)$
$\text{(1d)}$: integrate


Parametrize the torus as follows: at each point of the circle around the $z$-axis, $a(\cos(\phi),\sin(\phi),0)$ put a circle perpendicular to this circle: $$ \begin{align} p(\phi,\theta) &=\overbrace{a(\cos(\phi),\sin(\phi),0)}^\text{primary circle}+\overbrace{a(\cos(\phi)\cos(\theta),\sin(\phi)\cos(\theta),\sin(\theta))}^\text{secondary circle around the primary circle}\\ &=a(\cos(\phi)(1+\cos(\theta)),\sin(\phi)(1+\cos(\theta)),\sin(\theta))\tag{2a}]\\[6pt] p_1(\phi,\theta) &=a(-\sin(\phi)(1+\cos(\theta)),\cos(\phi)(1+\cos(\theta)),0)\\ &=a(1+\cos(\theta))(-\sin(\phi),\cos(\phi),0)\tag{2b}\\[6pt] p_2(\phi,\theta) &=a(-\cos(\phi)\sin(\theta),-\sin(\phi)\sin(\theta),\cos(\theta))\tag{2c} \end{align} $$ Thus, we get $$ \begin{align} |p_1(\phi,\theta)\times p_2(\phi,\theta)| &=a^2(1+\cos(\theta))\,|(\cos(\theta)\cos(\phi),\cos(\theta)\sin(\phi),\sin(\theta))|\\ &=a^2(1+\cos(\theta))\tag3 \end{align} $$ and we can compute $$ \int_0^{2\pi}\int_0^{2\pi}a^2(1+\cos(\theta))\,\mathrm{d}\phi\,\mathrm{d}\theta=4\pi^2a^2\tag4 $$ Theorem of Pappus

As I mentioned in a comment, we can apply the Theorem of Pappus: the primary circle has circumference $2\pi a$ and the secondary circle has circumference $2\pi a$, so the area is $$ (2\pi a)(2\pi a)=4\pi a^2\tag5 $$

  • $\begingroup$ The author defined the surface area by this: $$S=\int_a^b 2\pi y\sqrt{1+(\frac{dy}{dx})^2}dx$$ $\endgroup$
    – Doobius
    Jan 25 at 18:20
  • $\begingroup$ That would be for revolving about the $x$-axis. It would be useful to mention that. This would require two integrals, one for the part $y\le a$ and one for the part $y\ge a$, you can't simply double the one for the part $y\ge a$ (which has a larger surface area than the part $y\le a$). In my answer, I was revolving about the $y$-axis. With this new information, I will update the first part of my answer. $\endgroup$
    – robjohn
    Jan 25 at 18:30
  • 2
    $\begingroup$ Hmm... it was bothering me why symmetry consideration was not possible here. I didn't realize that the upper semicircle would indeed produce a greater surface area than the lower one until I read your comment. Thanks! $\endgroup$
    – Doobius
    Jan 25 at 18:35
  • 1
    $\begingroup$ You’re welcome. If you rotate around the $x$-axis, the upper and lower parts are the same, so there is a little less work. I’ve added, and expanded, the part I used to have in my answer when I thought you were rotating around the $x$-axis. $\endgroup$
    – robjohn
    Jan 26 at 20:44

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