I've been working on a homework set for Calc III, right now we're emphasizing double integration and polar integrals. I keep having problems conceptualizing where to actually create my region of integration (and subsequently the actual integrand)

Here is the most recent problem

$$ \text{Using polar coordinates, evaluate the integral}\\\text{which gives the area which lies in the first quadrant between the circles }\\ x^2 + y^2 = 196\text{ and } x^2 - 14x + y^2 = 0 $$

So I start by setting up my problem and I graph out the circles Area inscribed by two circles

I can see the area between the circles, and if I were to approach this in cartesian terms I would probably integrate y variable first from one equation to the second equation, and then the x variable from 0 to their intercept (which is probably 14 which I mentally deduced by the coordinates and radii, I haven't actually solved it). I also know I need to come up with something for the integrand, but I wouldn't even know what to do there.

So here's what my thinking is so far for polar coordinates. I believe that because it is just the change in r and the change in theta, I should be able to just do integrate the arclength like this:

$$ \int_{0}^{\frac{\pi}{4}} \int_{7}^{14}r\:dr\:d\theta\\ = \frac{147 \pi}{8} $$

From my inexperienced perspective, I should have the correct answer but my homework says it is not correct. What do I need to be doing differently?


2 Answers 2



The integral you are using gives the area of the region $\{(r,\theta):0\le\theta\le\pi/4,7\le r\le 14\}$. The region you are interested in is $\{(r,\theta):0\le\theta\le\pi/2,14\cos\theta\le r\le 14\}$.

  • $\begingroup$ This did yield the correct answer once I walked through all the integration. How do we get these endpoints? I don't understand that at all $\endgroup$ Jul 25, 2015 at 2:28
  • 1
    $\begingroup$ Circumference $x^2+y^2=196$ has center in $O$ and radius $14$ so $r=14$ is it equation in polar coordinates. Also, \begin{align*} x^2-14x+y^2&=0\\ \iff x^2+y^2&=14x \\ \iff r^2 & =14r\cos \theta \end{align*} From last equation we have $r=0$ or $r=14\cos \theta$ last one is the equation of the interior circumference in your picture. Then, the coordinates $(r,\theta)$ of the points of the region you are interested in satisfies the conditions $$0\le\theta\le \pi/2\qquad\qquad\text{and}\qquad\qquad 14\cos\theta \le r \le 14.$$ $\endgroup$ Jul 25, 2015 at 2:35
  • $\begingroup$ Thank you, I would not have come up with this answer $\endgroup$ Jul 25, 2015 at 3:04

You just made an error in one of your endpoints for integration, it should be:

$$ \int_{0}^{\frac{\pi}{4}} \int_{0}^{14}r\:dr\:d\theta\\ = \frac{49\pi}{2} $$

Think about it, why start half way into the smaller circle, we want the entire top half, not half of the top half.

Note: You technically can do this without Calculus just using the standard area formula for a circle and subtracting one half the area of the smaller circle from one fourth the area of the larger circle.

  • $\begingroup$ I'm picking this answer, because it is much easier for me to visualize and understand. I can see where I went wrong and it is clear that there is a change from $r$ starting at 0 (in the lower right corner) changing into 14 as it rotates about the origin to $\frac{\pi}{2}$ $\endgroup$ Jul 25, 2015 at 3:24

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