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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 $r = \cos\theta$.

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

I spend some time figuring this out...
At first I need find intersection (ie: $\sin\theta = \cos\theta$) between this 2 equations but this
obviously didn't made any sense.
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: overlay of both equation

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    The answer is actually (pi/8)- (1/4)2016-03-08
  • 0
    yes the area is $\frac{\pi}{8}-\frac{1}{4}$2016-03-14

3 Answers 3

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For the curve $r=\sin(\theta)$ $$ \left.\begin{array}{} x=r\cos(\theta)=\sin(\theta)\cos(\theta)\\ y=r\sin(\theta)=\sin^2(\theta) \end{array}\right\}\quad x^2+\left(y-\tfrac12\right)^2=\tfrac14 $$ For the curve $r=\cos(\theta)$ $$ \left.\begin{array}{} x=r\cos(\theta)=\cos^2(\theta)\\ y=r\sin(\theta)=\sin(\theta)\cos(\theta) \end{array}\right\}\quad\left(x-\tfrac12\right)^2+y^2=\tfrac14 $$ So the area is the intersection of the two circles:

enter image description here

The area of the purple lune is $\frac14$ of the area of a radius $\frac12$ circle minus the area of a $\frac12,\frac12,\frac1{\sqrt2}$ right triangle, that is $$ \frac14\cdot\frac\pi4-\frac12\cdot\frac12\cdot\frac12 $$ Since the area of the green lune is the same, the area of the given intersection is $$ \frac\pi8-\frac14 $$

5

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.

enter image description here

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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.2012-07-17
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    @kypronite: Yes I saw that. I hope polar plot help you. :)2012-07-17
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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.

  • 0
    The answer I get with [Wolframalpha](http://www.wolframalpha.com/input/?i=integral+from+0+to+Pi%2F4+of+%28sin%28x%29%29%5E2) is not quite what you have though.2012-07-17
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    shouldn't the integrand be $(\sin\theta - \cos\theta)^2$2012-07-17
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    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.2012-07-17
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    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...2012-07-17
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    Yes, that's what I compute.2012-07-17
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    thanks.Cause it seem strange, I think i'm confusing unit circle and polar coordinate.Looks like I have more reading to do on polar coordinate.2012-07-17
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    As a control, you can also compute it using [the technique proposed here](http://math.stackexchange.com/questions/171856/area-of-shaded-region/171860#171860).2012-07-17