# Area of a Polar Curve Above the Polar Axis

• May 14th 2009, 12:36 PM
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
• May 14th 2009, 01:13 PM
galactus
Is this what you're looking for?.

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

This gives the area from 0 to Pi. In other words, the smaller portion above the 'x-axis', so to speak.
• May 14th 2009, 01:22 PM
yeah. i did it from 0 to pi/2 though...why is it pi?
• May 14th 2009, 01:33 PM
galactus
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.
• May 14th 2009, 01:36 PM
Soroban

Did you make a sketch?

Quote:

Find the area of the interior of $\displaystyle r\:=\:1-\sin\theta$ (above the polar axis)
This is a cardioid . . .
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: .$\displaystyle \frac{3\pi}{4}-2 \:\approx\:0.3562$

• May 14th 2009, 01:38 PM
galactus
Quote:

Originally Posted by Soroban

Did you make a sketch?

This is a cardioid . . .
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: .$\displaystyle \frac{3\pi}{4}-2 \:\approx\:0.3562$

There ya' go. I just did it from 0 to Pi. Gives the same thing.
• May 14th 2009, 01:41 PM