Thread: trying to find area in polar form

I just have a quick question about setting up the problem.
Given r=1-1sin(theta)
how do you find the angles to use for the upper and lower limits of the integral. I have been setting r=0 but it still requires some finesse, there has to be an easier or more proper way.

thanks in advance.

Originally Posted by calculushelptx

I just have a quick question about setting up the problem.
Given r=1-1sin(theta)
how do you find the angles to use for the upper and lower limits of the integral. I have been setting r=0 but it still requires some finesse, there has to be an easier or more proper way.

thanks in advance.

What area are you trying to find?

Find the area of the region lying between the inner and outer loops of the limacon r=1-2sin(theta). I know that you just set up an integration but I cannot figure out how you get the upper and lower limits.

If you are trying to find the area in general you look at the interval of theta which generates the graph. here you have a cardioid so theta varies from 0 to 2pi.

if you have 2 graphs such as the cardioid r1 1+ cos(t) and the circle r = 3/2
and want the area inside the cardioid but outside the circle then
you would set r1 = r to find the limits of integration (here -pi/3 to pi/3)

Here is a better example.

How do I find the radian angles of cos2(theta)=r to find the area of one petal.

For for r = 1-2sin(t)

knowing the graph is the whole story

As with any polar graph use the usual suspects 0 - pi/2-pi 3pi/2 and 2pi

and add the zeroes here pi/6 and 5pi/6 this gives you just the maxs, mins and zeroes which is all you need


Half the outer loop is completed as theta varies from -p1/2 to pi/6-Multiply this integral by 2

Half The inner loop is generated as theta varies pi/6 to pi/2 multiply by 2 and subtract this from previous

Its all knowing what to look for

For cos(nt) and sin(nt) consider pi/2n Again these give the zeroes maxs and mins

here pi/4. you get one half petal as t varies from 0 to pi/4