# Thread: Area between two curves in polar coords

1. ## Area between two curves in polar coords

Well this disturbs me.

We have two intersecting curves
$r = \sqrt{3}~cos(\theta)$
and
$r = sin(\theta)$

We want the area between the curves.

Now, according to my text the area contained only within the cosine curve will be
$\frac{1}{2} \int_0^{\pi / 3} (3~cos^2(\theta) - sin^2(\theta) ) ~d \theta$

So then we subtract this from the area of the cosine curve.
$A = \pi \left ( \frac{\sqrt{3}}{2} \right ) ^2 - \frac{1}{2} \int_0^{\pi / 3} (3~cos^2(\theta) - sin^2(\theta) ) ~d \theta$

The student was insisting that the answer is this:
$A = \frac{1}{2} \int_0^{\pi / 3} sin^2(\theta)~d\theta + \frac{1}{2} \int_{\pi / 2}^{\pi / 3} sin^2(\theta)~d\theta$

I know that the text is right (if I set it up properly, that is), but the region of integration of the cosine circle doesn't, in fact, start with the angle 0.

It's been a long time since I've done these, folks. Please forgive me if I'm missing the obvious.

-Dan

2. Originally Posted by topsquark
Well this disturbs me.

We have two intersecting curves
$r = \sqrt{3}~cos(\theta)$
and
$r = sin(\theta)$

We want the area between the curves.

Now, according to my text the area contained only within the cosine curve will be
$\frac{1}{2} \int_0^{\pi / 3} (3~cos^2(\theta) - sin^2(\theta) ) ~d \theta$

So then we subtract this from the area of the cosine curve.
$A = \pi \left ( \frac{\sqrt{3}}{2} \right ) ^2 - \frac{1}{2} \int_0^{\pi / 3} (3~cos^2(\theta) - sin^2(\theta) ) ~d \theta$

The student was insisting that the answer is this:
$A = \frac{1}{2} \int_0^{\pi / 3} sin^2(\theta)~d\theta + \frac{1}{2} \int_{\pi / 2}^{\pi / 3} sin^2(\theta)~d\theta$

I know that the text is right (if I set it up properly, that is), but the region of integration of the cosine circle doesn't, in fact, start with the angle 0.

It's been a long time since I've done these, folks. Please forgive me if I'm missing the obvious.

-Dan
Your student is almost correct as far as I see. I think it's:

$A = \frac{1}{2} \int_0^{\pi/3} \sin^2 \theta \, d \theta + \frac{1}{2} \int_{{\color{red}\pi/3}}^{{\color{red}\pi/2}} 3 \cos^2 \theta \, d\theta$.

The first integral is the (smaller) area between the rays $\theta = 0$, $\theta = \frac{\pi}{3}$ and the sine circle. The second integral is the (smaller) area between the rays $\theta = \frac{\pi}{3}$, $\theta = \frac{\pi}{2}$ and the cosine circle.

I get $\frac{5 \pi - 6 \sqrt{3}}{24}$ as the answer (the book's answer too, I hope).

Note that the point (0, 0) lies on the sine circle at [0, 0] and at [0, pi/2] for the cosine circle.

So the problem, I think (and as you seem to realise), with how you're setting it up is that the polar coordinates for the point (0, 0) are different for the two curves so you can't use that difference integral .....

3. Originally Posted by mr fantastic
Your student is almost correct as far as I see. I think it's:

$A = \frac{1}{2} \int_0^{\pi/3} \sin^2 \theta \, d \theta + \frac{1}{2} \int_{{\color{red}\pi/3}}^{{\color{red}\pi/2}} 3 \cos^2 \theta \, d\theta$.

The first integral is the (smaller) area between the rays $\theta = 0$, $\theta = \frac{\pi}{3}$ and the sine circle. The second integral is the (smaller) area between the rays $\theta = \frac{\pi}{3}$, $\theta = \frac{\pi}{2}$ and the cosine circle.

I get $\frac{5 \pi - 6 \sqrt{3}}{24}$ as the answer (the book's answer too, I hope).

Note that the point (0, 0) lies on the sine circle at [0, 0] and at [0, pi/2] for the cosine circle.

So the problem, I think (and as you seem to realise), with how you're setting it up is that the polar coordinates for the point (0, 0) are different for the two curves so you can't use that difference integral .....
Okay, so its more or less a problem with the translation of $\theta$ in the cosine integral.

I'm simply not able (yet) to get a picture of why this happens is my main problem right now. I mean, this kind of thing doesn't occur when we to this problem in Cartesian coordinates. (Hmmm, that's a thought. Maybe I'll set it up that way and see what happens.)

-Dan

4. Originally Posted by topsquark
Okay, so its more or less a problem with the translation of $\theta$ in the cosine integral.

I'm simply not able (yet) to get a picture of why this happens is my main problem right now. I mean, this kind of thing doesn't occur when we to this problem in Cartesian coordinates. (Hmmm, that's a thought. Maybe I'll set it up that way and see what happens.)

-Dan
You get the same answer when adding the areas of the two segments together:

The cartesian coordinates of the two intersection points are (0, 0) and $\left( \frac{\sqrt{3}}{4}, \, \frac{3}{4} \right)$.

Therefore the length of the chord of the segment in each circle is $\frac{\sqrt{3}}{2}$.

Radius of sine circle is $\frac{1}{2}$. Apply cosine rule to the triangle formed by the chord of the segment and the two radii to find that the angle subtended at the centre by the segment is $\frac{2 \pi}{3}$.

Radius of cosine circle is $\frac{\sqrt{3}}{2}$. Therefore the chord of the segment and the two radii form an equilateral triangle. Therefore the angle subtended at the centre by the segment is $\frac{\pi}{3}$.

Area of sine circle segment = $\frac{1}{2} \left( \frac{1}{2} \right)^2 \left( \frac{2 \pi}{3} - \sin \frac{2 \pi}{3} \right) = \frac{1}{8} \left( \frac{2 \pi}{3} - \frac{\sqrt{3}}{2} \right) = \frac{\pi}{12} - \frac{\sqrt{3}}{16}$.

Area of cosine circle segment = $\frac{1}{2} \left( \frac{\sqrt{3}}{2} \right)^2 \left( \frac{\pi}{3} - \sin \frac{\pi}{3} \right) = \frac{3}{8} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) = \frac{3\pi}{24} - \frac{3\sqrt{3}}{16}$.

Area between the two circles = (area of sine circle segment) + (area of cosine circle segment)

$= \left( \frac{\pi}{12} - \frac{\sqrt{3}}{16} \right) + \left( \frac{3\pi}{24} - \frac{3\sqrt{3}}{16} \right) = \frac{5 \pi}{24} - \frac{4 \sqrt{3}}{16} = \frac{5 \pi}{24} - \frac{6 \sqrt{3}}{24}$.