# Math Help - Area of a Polar Curve Above the Polar Axis

1. ## Area of a Polar Curve Above the Polar Axis

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

2. Is this what you're looking for?.

$\frac{1}{2}\int_{0}^{\pi}(1-sin(t))^{2}dt=\frac{3\pi-8}{4}$

This gives the area from 0 to Pi. In other words, the smaller portion above the 'x-axis', so to speak.

3. yeah. i did it from 0 to pi/2 though...why is it pi?

4. 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.

Did you make a sketch?

Find the area of the interior of $r\:=\:1-\sin\theta$ (above the polar axis)
This is a cardioid . . .
Code:
                |
..*..   |   ..*..
*:::::::* | *:::::::*
- - * - - - - - * - - - - - * - -
*            |            *
|
*             |             *
*             |             *
*             |             *
|
*            |            *
*           |           *
*         |         *
*      |      *
* * *
|

We can find the area from $\theta = 0$ to $\theta = \tfrac{\pi}{2}$ . . . and double.

So we have:. $A \;=\;2 \times \tfrac{1}{2}\int^{\frac{\pi}{2}}_0(1 -\sin\theta)^2\,d\theta$

I get: . $\frac{3\pi}{4}-2 \:\approx\:0.3562$

6. Originally Posted by Soroban

Did you make a sketch?

This is a cardioid . . .
Code:
                |
..*..   |   ..*..
*:::::::* | *:::::::*
- - * - - - - - * - - - - - * - -
*            |            *
|
*             |             *
*             |             *
*             |             *
|
*            |            *
*           |           *
*         |         *
*      |      *
* * *
|

We can find the area from $\theta = 0$ to $\theta = \tfrac{\pi}{2}$ . . . and double.

So we have:. $A \;=\;2 \times \tfrac{1}{2}\int^{\frac{\pi}{2}}_0(1 -\sin\theta)^2\,d\theta$

I get: . $\frac{3\pi}{4}-2 \:\approx\:0.3562$

There ya' go. I just did it from 0 to Pi. Gives the same thing.

7. thank you both very much