# Math Help - find the area shaded by the circles

1. ## find the area shaded by the circles

find the area shared by the circles r=2cos(theta) and r=2sin(theta).

i know the general formula for finding the area, but i don't know which is the outside one, and which is the inside circle. but i've tried both ways and am still not getting the right answer.

so let's just say i'll try it with integral .5(4cos(x)^2-.5(4sin(x)^2)
i can factor out a 2, giving me cos(x)^2-sin(x)^2. the integral of that, i believe, can be expressed as 4sin(2x). now i figured the limits of integration were from 0 to pi/4, because those are the two places the circles intersect. so evaluating there, i would get 4-0=4. but i've been told the answer is pi/2 -1. so where am i going wrong, because i'm way off.

2. Originally Posted by isuckatcalc
find the area shared by the circles r=2cos(theta) and r=2sin(theta).

i know the general formula for finding the area, but i don't know which is the outside one, and which is the inside circle. but i've tried both ways and am still not getting the right answer.

so let's just say i'll try it with integral .5(4cos(x)^2-.5(4sin(x)^2)
i can factor out a 2, giving me cos(x)^2-sin(x)^2. the integral of that, i believe, can be expressed as 4sin(2x). now i figured the limits of integration were from 0 to pi/4, because those are the two places the circles intersect. so evaluating there, i would get 4-0=4. but i've been told the answer is pi/2 -1. so where am i going wrong, because i'm way off.
use symmetry ...

$A = 2 \int_0^{\frac{\pi}{4}} \frac{(2\sin{t})^2}{2} \, dt
$

$A = 4 \int_0^{\frac{\pi}{4}} \sin^2{t} \, dt$

$A = 4 \int_0^{\frac{\pi}{4}} \frac{1-\cos(2t)}{2} \, dt$

$A = 2 \int_0^{\frac{\pi}{4}} 1-\cos(2t) \, dt$

finish

3. ok i get it now, i was trying to use both r=2cos(x) and r=2sin(x). how does one know when to use only one or when to subtract one from the other? since there were two graphs it seemed obvious to me that the latter formula needed to be used, but apparently it didn't. also, how do you know to go with r=2sin(x) instead of r=2cos(x)? thanks for the help.

4. Originally Posted by isuckatcalc
ok i get it now, i was trying to use both r=2cos(x) and r=2sin(x). how does one know when to use only one or when to subtract one from the other? since there were two graphs it seemed obvious to me that the latter formula needed to be used, but apparently it didn't. also, how do you know to go with r=2sin(x) instead of r=2cos(x)? thanks for the help.
You have to look at the graph and the region area you want to find. LOOK for symmetry and take advantage of it.

I could have went with cosine ...

$A = 2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{(2\cos{\theta})^2}{2} \, d\theta$

5. Converting to Cartesian coordinates, $x= r cos(\theta)$ and $y= r sin(\theta)$ so $r= 2cos(\theta)$ or $r^2= 2 rcos(\theta)$ and so $x^2+ y^2= 2 x$. Then $x^2- 2x+ y^2= 0$ and, completing the square, $x^2- 2x+ 1+ y^2= (x- 1)^2+ y^2= 1$. That is the circle with center at (1, 0) and radius 1. It is tangent to the y-axis at the origin.

Similarly, $r= 2 sin(\theta)$ becomes $r^2= 2 r sin(\theta)$ or $x^2+ y^2= y$ giving the circle with center at (0,1) and radius 1. It is tangent to the x-axis at the origin. There is no "inside" or "outside". They overlap on the first quadrant, having x=y or $\theta= \pi/4$ as the mid line. That should lead you to integrate $r= 2 sin(\theta)$ from $\theta= 0$ to $\pi/4$ and then double to get the entire area.