Polar Coordinate Area between two curves

cryora

1. The problem statement, all variables and given/known data
Find the area enclosed by the curves:
[tex]r=\sqrt(3)cos(\theta)[/tex]
and
[tex]r=sin(\theta)[/tex]

2. Relevant equations
The area between two polar curves is given by:
[tex]A=(1/2)\int{R^2 - r^2dr}[/tex] where R is the larger function and r is the smaller function over an interval.

3. The attempt at a solution
I set:
[tex]\sqrt{3}cos(\theta)=sin(\theta)[/tex]
[tex]\sqrt(3)=tan(\theta)[/tex]
[tex]\theta=\pi/3,4\pi/3[/tex]

Graphically, I can see that when [tex]0<\theta<\pi/3[/tex] or [tex]4\pi/3<\theta<\pi[/tex] that [tex]sin(\theta)<\sqrt{3}cos(\theta)[/tex]
and when [tex]\pi/2<\theta<4\pi/3[/tex], [tex]\sqrt{3}cos(\theta)<sin(\theta)[/tex]
So it follows that I will have:
[tex]A=(1/2)\int(0,\pi/3){3cos^2(\theta)-sin^(\theta)d\theta+(1/2)\int(\pi/3,4\pi/3){sin^(\theta)-3cos^2(\theta)d\theta+(1/2)\int(4\pi/3,2\pi){3cos^2(\theta)-sin^(\theta)d\theta[/tex]
The numbers separated by commas inside the parenthesis are the limits of integration. Sorry I'm new at this.

I'm just wondering if this is the right way to set it up for a question like this.

LCKurtz

What you have isn't even close to what you need to do. The first thing you need to do is draw a polar coordinate picture of your two curves and identify visually what area is between the curves. Do you know what the graphs look like?

cryora

Ok, I realize my foolish mistake. So it appears to be two circles, with intersections at (√(3)/2, ∏/3) and the pole. I guess what I do now is integrate once from 0 to ∏/3 using r = sin(θ), and add another integral from ∏/3 to ∏/2 using r = √(3)cos(θ).

So what I'll have is:
[tex]\frac{1}{2}\int_0^\frac{\pi}{3} sin^2(\theta) \,d\theta + \frac{1}{2}\int_\frac{\pi}{3}^\frac{\pi}{2} 3cos^2(\theta) \,d\theta[/tex]

I hope this is correct?

LCKurtz

Much better. Amazing how a picture helps, eh?

flyingpig

I am actually confused by this question



For his second integral

[tex]\frac{1}{2}\int_\frac{\pi}{3}^\frac{\pi}{2} 3cos^2(\theta) \,d\theta[/tex]

Why is it from pi/3 to pi/2? I don't see it.

SammyS

Because the larger circle is tangent to the y-axis .

flyingpig

Oh okay never mind i see it. pi/3 to pi/2 sweeps half of the loop and from 0 to pi/3 sweeps the other half