The picture below is the solution to the following problem as presented in my book:

Find the area of the region that lies inside both curves $$r = 8 + \cos \theta \\r = 8 − \cos θ$$

enter image description here

According to my book, it is possible to solve the problem in such a way because of "symmetry" but does not explain what "symmetry" means. From my basic understanding of the meaning of the word in every day usage, I can see that a reflection over the vertical axis, in this case, does not affect the shape of the graph, and I can also intuitively see that $$A = 4\int_0^\frac{\pi}{2}\frac{1}{2}(6-\cos\theta)^2d\theta$$ This, as it turns out, is true but is not satisfying because it was an intuitive guess. So, what is the mathematical definition of symmetry and how can one show that this graph is symmetric? i.e. how to explain mathematically that this problem could be solved using "symmetry"?

  • $\begingroup$ I imagine you do not really mean $8+\cos\theta$ and $8-\cos\theta$. The symmetry is just the fact that the shapes in the $4$ quadrants are the same. And as quoted, symmetry is not even taken full advantage of. There is a subtler "symmetry" $0$ to $\pi/4$ and $\pi/4$ to $\pi/2$. $\endgroup$ May 2, 2013 at 3:47

3 Answers 3


Purely for fun, let's break up the integral from $0$ to $\pi/2$ into two integrals, $0$ to $\pi/4$ and $\pi/4$ to $\pi/2$. For the second integral, make the change of variable $\phi=\pi/2-\theta$. Note that $\cos\theta=\sin \phi$. So our integral becomes $$\int_0^{\pi/4}\frac{1}{2}(6-\cos\theta)^2\,d\theta + \int_0^{\pi/4}\frac{1}{2}(6-\sin\phi)^2\,d\phi.$$ Change the second variable back to $\theta$. We get that our area is $$\frac{1}{2}\int_0^{\pi/4}\left[(6-\cos\theta)^2+(6-\sin\theta)^2\right]\,d\theta.$$ Expand, integrate, noting that $\sin^2\theta+\cos^2\theta=1$. An antiderivative is $$\frac{1}{2}\left[73\theta -12(\sin\theta-\cos\theta)\right].$$ "Plug in," noting that at $0$ we have $\sin\theta-\cos\theta=-1$. We end up with $\frac{73\pi}{8} -6$. Then multiply by $4$.

We took advantage of a polar "symmetry" about $r=6$. The bonus is that we didn't have to integrate $\cos^2\theta$. It is not clear whether it was worth it.


The mathematical definition of symmetry is really given through group theory, but you really don't need any sort of group-theoretic analysis here. A symmetry of something is a transformation that leaves it unchanged. For example, $x^2$ has the symmetry $x\to -x$, because $(-x)^2=x^2$. Making this transformation left our equation unchanged, so the equation has this symmetry. Functions with this symmetry ($x\to-x$) are called even functions. The expressions $r=6\pm\cos(\theta)$ are even, so they also have this particular symmetry. Replacing $\theta$ with $-\theta$ leaves the equation intact, which gives us the symmetry across the $x$-axis. The vertical symmetry is somewhat less obvious. To swap everything across the $y$-axis, we do $\theta\to\pi-\theta$.


If you know your trig identities, this comes to


So what this symmetry does is swap the equations: each equation doesn't have this symmetry by itself, but the system of equations does, because it turns each curve into the other. These two conditions together tell you that you can treat each of the quadrants on equal grounds, and so you only need to consider one to get your answer.

Seeing which symmetries an equation has is not generally obvious. It's easier just to have a few rules of thumb to check: symmetry across the axes, even/odd functions, symmetry made by swapping variables, and that sort of thing.


Symmetry means what you think it means -- saying that a shape is symmetric about the line $L$ means that reflecting through $L$ gives you back the same shape you started with. Or, saying it another way, the part of the shape on one side of the line $L$ looks the same as the part on the other side. So, in your diagram, each individual curve is symmetric about the $x$-axis. More importantly, the area we're analyzing is symmetric about both the $x$-axis and the $y$-axis.

I could give you a more formal definition of symmetry, full of jargon and symbols, but I don't think it would help, in this situation.

So, "by symmetry", you can calculate the area in the top-right quadrant, and multiply that answer by 4 to get the total area. The top-right quadrant corresponds to $0 \le \theta \le \tfrac{\pi}2$, so that's the interval over which you integrate.

The integrand $\tfrac{1}{2}(6-\cos\theta)^2$ is correct only if the original curve equations are $r = 6 \pm \cos\theta$. It looks like somewhere the number "6" got switched to "8".

We're using $r = 6 - \cos\theta$ in our integrand because this is the curve that borders the shaded region in the top-right quadrant. The curve $r = 6 - \cos\theta$ is "inside" the curve $r = 6 + \cos\theta$ in the top-right quadrant because $6 - \cos\theta \le 6 + \cos\theta$ when $0 \le \theta \le \tfrac{\pi}2$. No guessing required :-).


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