# area between polar equation $r = \sin\theta$ and $r = \cos\theta$

Below is the exact question and answer from my textbook:
Find the area of the region enclosed between the two curves $C_{1}$ and $C_{2}$ where $C_{1}$ has the polar equation $r = \sin\theta$ and $C_{2}$ has the polar equation $r = \cos\theta$.

$\frac{\pi}{8} - \frac{1}{16}$

I spend some time figuring this out...
At first I need find intersection (ie: $\sin\theta = \cos\theta$) between this 2 equations but this
But then how would i find lower and upper limit [a,b]
using the formula for area = $\int_a^b \frac{1}{2}(\sin\theta-\cos\theta)^2 \,d\theta$

i assume the textbook is asking area for this:

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The sine and the cosine are equal when $\theta=\pi/4$. The two curves are actually circles with radii 1/2 and center $(0,1/2)$ and $(1/2,0)$ for the $\sin$ and $\cos$ respectively. You can thus find the area by computing the following integral

$$\int_0^{\pi/4} (\sin\theta)^2 d\theta$$

in which I multiplied by $2$ exploiting symmetry.

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The answer I get with Wolframalpha is not quite what you have though. –  Raskolnikov Jul 17 '12 at 14:22
shouldn't the integrand be $(\sin\theta - \cos\theta)^2$ –  kypronite Jul 17 '12 at 14:42
No, I use the polar integral $\frac{1}{2}\int r^2 d\theta$. And I only integrate over half of the surface, which means I only need the curve $r=\sin\theta$. I then double to recover the full surface area. –  Raskolnikov Jul 17 '12 at 14:47
I just edited my post to include image of overlay of both polar equations.Can you confirm that's what actually your solution give?Sorry,I'm still learning multivariable calculus... –  kypronite Jul 17 '12 at 15:05
Yes, that's what I compute. –  Raskolnikov Jul 17 '12 at 15:06

I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.

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i already uploaded polar plot same time as yours.I think i try to visualize polar coordinate as unit circle, hence the confusion in my part. –  kypronite Jul 17 '12 at 15:12
@kypronite: Yes I saw that. I hope polar plot help you. :) –  Babak S. Jul 17 '12 at 15:14
very nice plot! +1 –  amWhy Mar 12 '13 at 1:49