find the area of the interior of r=1-sin theta
(above the polar axis)
i think it is pi/4 but i also got pi/8 and i'm not sure which one, or if either, is right... thank you in advance to anyone who helps
I just integrated from 0 to Pi. You can do it from 0 to Pi/2 and multiply by 2 as well to get the same thing.
The problem said area above the polar axis. It starts at what we can say is the positive x-axis and turns counterclockwise to Pi so that it encompasses all the area above the polar axis. After you pass Pi(180 degrees) you get below it. I shaded the wanted region, but it did not appear when I saved it to post. Sorry 'bout that.
Hello, andyaddition!
Did you make a sketch?
This is a cardioid . . .Find the area of the interior of $\displaystyle r\:=\:1-\sin\theta $ (above the polar axis)Code:| ..*.. | ..*.. *:::::::* | *:::::::* - - * - - - - - * - - - - - * - - * | * | * | * * | * * | * | * | * * | * * | * * | * * * * |
We can find the area from $\displaystyle \theta = 0$ to $\displaystyle \theta = \tfrac{\pi}{2}$ . . . and double.
So we have:.$\displaystyle A \;=\;2 \times \tfrac{1}{2}\int^{\frac{\pi}{2}}_0(1 -\sin\theta)^2\,d\theta $
I get: .$\displaystyle \frac{3\pi}{4}-2 \:\approx\:0.3562$