# Finding the area of a cardioid

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• Feb 12th 2007, 03:23 PM
pakman
Finding the area of a cardioid
The problem states that S is the region inside the cardioid r = 6 - 6sin(theta)

It says we are to solve the problem by calculating Double integral r dr dtheta

I'm fairly comfortable solving double integrals, but I've forgotten how to find the bounds. Thanks in advance.
• Feb 12th 2007, 03:33 PM
pakman
Here's even another problem which we did in class. Assume S = integral

evaluate this integral SS(4 - x^2 - y^2)dxdy. R is the first quadrant sector of the circle x^2 + y^2 = 4 between the lines x=0 and y=x

0 <= r <= 2 and 0 <= theta <= pi/4

Where in the world did my teacher get these bounds?
• Feb 12th 2007, 04:00 PM
ThePerfectHacker
Quote:

Originally Posted by pakman
The problem states that S is the region inside the cardioid r = 6 - 6sin(theta)

It says we are to solve the problem by calculating Double integral r dr dtheta

I'm fairly comfortable solving double integrals, but I've forgotten how to find the bounds. Thanks in advance.

There is no "inner" radius thus it is zero.
The "outer" radius is the cardiod: r= 6-6sin(t)

It makes a complete revolution, hence from 0 to 2*pi.

Thus,
INT (from 0 to 2*pi) INT (from 0 to 6-6*sin(t)) csc t r dr dt

The reason why csc(t) is because you have 1 dy dx but 1 expressed in polar coordinates is csc(t). Because,
y=1
r*sin(t)=1
r=csc(t)