I am asked to find the portion of the area $r=2a \cos^2 (\frac{\theta}{2})$ and $r=2a \sin^2 (\frac{\theta}{2})$ have in common with each other provided $a>0$ on the polar plane.

Can someone help me set up the integral please? I'm having some trouble with that part.

I know the formula is:

enter image description here

If I set $r=2a \cos^2 (\frac{\theta}{2})$ and $r=2a \sin^2 (\frac{\theta}{2})$ equal to each other, I get

$2a\sin^2 (\frac{\theta}{2})=2a \cos^2 (\frac{\theta}{2})$


If I solve this, I get $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

How would I set up the integral now though? Like, which area would I subtract so I capture the entire domain of intersection? I am a little confused about that. Any help would be much appreciated! Thanks!


Let me give names to the two functions you mention. Let \begin{align} r_1(\theta) &= 2a \cos^2(\theta/2) \\ r_2(\theta) &= 2a \sin^2(\theta/2) \end{align} Based on this picture of $\color{green}{r_1}$ and $\color{red}{r_2}$ for $a=1$: Polar plot

It looks like you can compute the intersecting area by $$ A = \int_{-\pi/2}^{\pi/2} \frac{1}{2} \color{red}{r_2}(\theta)^2 \, d\theta + \int_{\pi/2}^{3\pi/2} \frac{1}{2} \color{green}{r_1}(\theta)^2 \, d\theta $$ But based on the symmetry, if you want to take a shortcut, you can probably just do: $$ A = 4 \int_0^{\pi/2} \frac{1}{2} \color{red}{r_2}(\theta)^2 \, d\theta = 4 \int_{\pi/2}^{\pi} \frac{1}{2} \color{green}{r_1}(\theta)^2 \, d\theta $$

  • $\begingroup$ How did you get those bounds? That's what I'm having trouble seeing. $\endgroup$ – Future Math person Mar 23 '17 at 23:14
  • $\begingroup$ @SubhashisChakraborty: Thanks. Fixed. And now for an explanation of the bounds... $\endgroup$ – WB-man Mar 23 '17 at 23:23
  • $\begingroup$ @SubhashisChakraborty: I wish I could make a GIF animation, that's what this answer really needs, but I'll try to explain with text: Consider $r_2$. Imagine I drew a line coming out from the origin in the direction $\theta$ that stops once it hits the $r_2$ curve. If I start from $\theta=-\pi/2$ and then sweep this line counter-clockwise to $\theta=\pi/2$, you'll see that it sweeps the right half of the intersection region. Likewise, if you do the same for the $r_1$ curve, you'll see that sweeping from $\theta=\pi/2$ to $3\pi/2$ gets the left half. $\endgroup$ – WB-man Mar 23 '17 at 23:27
  • $\begingroup$ That makes a lot of sense actually. I just need to see WHERE $\frac{\pi}{2}$ is on this polar plot. That's where I am having trouble. Where are all these points of intersections on this plot? I was under the assumption we start at 0, since that is a point of intersection. $\endgroup$ – Future Math person Mar 23 '17 at 23:31
  • $\begingroup$ @SubhashisChakraborty: Remember $\pi/2$ corresponds to $90^\circ$. These angles are always measured by going counter-clockwise from the postive $x$-axis. Hence $\theta=\pi/2$ is the positive $y$-direction, $-\pi/2$ is the negative $y$-direction (going a quarter turn clockwise) and so is $3\pi/2$ (three quarter turns counter-clockwise). $\endgroup$ – WB-man Mar 23 '17 at 23:34

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