Polar Coordinate Area between two curves

[cryora]

Homework Statement

Find the area enclosed by the curves:
r=\sqrt(3)cos(\theta)
and
r=sin(\theta)

Homework Equations

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

The Attempt at a Solution

I set:
\sqrt{3}cos(\theta)=sin(\theta)
\sqrt(3)=tan(\theta)
\theta=\pi/3,4\pi/3

Graphically, I can see that when 0<\theta<\pi/3 or 4\pi/3<\theta<\pi that sin(\theta)<\sqrt{3}cos(\theta)
and when \pi/2<\theta<4\pi/3, \sqrt{3}cos(\theta)<sin(\theta)
So it follows that I will have:
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
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:
\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

I hope this is correct?

 

[LCKurtz]

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:
\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

I hope this is correct?

Much better. Amazing how a picture helps, eh?

 

[flyingpig]

I am actually confused by this question

[PLAIN]http://img3.imageshack.us/img3/3046/unledze.jpg

For his second integral

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

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

 

[SammyS]

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

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