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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

  • 0
    yes the area is $\frac{\pi}{8}-\frac{1}{4}$2016-03-14

3 Answers 3

1

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 $

6

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

  • 0
    @kypronite: Yes I saw that. I hope polar plot help you. :)2012-07-17
3

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
    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