- #1
cryora
- 51
- 3
Homework Statement
Find the area enclosed by the curves:
[tex]r=\sqrt(3)cos(\theta)[/tex]
and
[tex]r=sin(\theta)[/tex]
Homework 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.
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.