# Finding AREA in POLAR COORDINATE

muffintop

## Homework Statement

Find the area inside one leaf of the four-leaved rose r = cos2x

## Homework Equations

A = 1/2 antiderivative abr2 dx

## The Attempt at a Solution

I just need help in finding the lower and upper limits of integration. But besides that, I know how to do the rest.
If my integration is right
A = 1/2 antiderivative cos2 2x
= 1/2 antiderivative (1 + cos 4x)/2
= 1/2 (1/2 x + 2 sin 4x​

It may help to visualize what is going on. The integral is adding many little wedges to give an area. Each wedge is a thin triangle, with the sharp point at the origin, distance "r" along the long edge, and a base of "dx" (a small angle) times r.

You want to add up all the wedges inside one rosette. So pick a value of "x" where "r" is zero. Let x increase from there, and r increase with it as you follow along the rosette. Keep going, until r comes back to zero again. You've now mapped out a single rosette. Your starting and ending values for x will be the bounds of where you want to add up all those wedges; and hence they are the bounds of the definite integral.

Cheers -- sylas

PS. Check your antiderivative, by getting its derivative again to see if you get what you started with.

muffintop
I'm sorry but I don't get it.
Can you show me, perhaps with a different example so it's not like you're doing the homework for me?

Gold Member
First of all, I'm assuming r=cos(2theta) is on [0,2pi].

You need to find an interval of theta such that for r=0. This can be done by solving 0=cos(2theta) for theta. This will generate several answers. You will need to figure out which solutions will give you an interval of theta for which one "petal" of the rose curve is generated. Try graphing the polar equation and then find values of r for intermediate values between the various answers to the 0=cos(2theta) equation.

We'll call that interval [a,b], for which when values from a to b are input into the polar equation one complete petal is traced. Then the problem is simple: just take the polar area integral dA=.5 int_a^b r^2 dtheta of r=cos(2theta) on that interval.