# Thread: Finding the area of a cardioid

1. ## 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.

2. 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?

3. 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)

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### how to calculate area of cardiod

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