Hit A Snag When Finding Area Inside A Circle And Under a Line

[DCBaelar]

Homework Statement

Find the total area inside the circle r = 4 and below the line r=2csc\theta

Homework Equations

\int^{b}_{a} 1/2r^{2}\thetad\theta

The Attempt at a Solution

r=2/sin\theta\Rightarrowrsin\theta=2\Rightarrowy=2
r=4\Rightarrow=circle with radius 4 at center (0,0)

Point of Intersection:
4=2/sin\theta\Rightarrow4sin\theta=2\Rightarrowsin\theta=2/4\Rightarrow\theta=\pi/6

\int^{pi/6}_{0} 1/2 (4-2csc\theta)^{2}d\theta
\int^{pi/6}_{0} 1/2 (16-4csc^{2}\thetad\theta)
\int^{pi/6}_{0} (8-2csc^{2}\thetad\theta)
(8\theta+2cot\theta)|^{pi/6}_{0}
(8*pi/6)+2cot(pi/6)-(8*0)-2cot(0)
4pi/3+2√3-0-undefined

And that's my problem..the undefined 2cot.

I think where I went wrong is that sinθ=1/2 at \pi/6 and 5\pi/6 and thus my boundaries of integration should be \pi/6 and 5\pi/6.

Am I on the right track?

Thanks for any feedback/assistance.

Jason

 

[LCKurtz]

Homework Statement

Find the total area inside the circle r = 4 and below the line r=2csc\theta

Homework Equations

\int^{b}_{a} 1/2r^{2}\thetad\theta

The Attempt at a Solution

r=2/sin\theta\Rightarrowrsin\theta=2\Rightarrowy=2
r=4\Rightarrow=circle with radius 4 at center (0,0)

Point of Intersection:
4=2/sin\theta\Rightarrow4sin\theta=2\Rightarrowsin\theta=2/4\Rightarrow\theta=\pi/6

\int^{pi/6}_{0} 1/2 (4-2csc\theta)^{2}d\theta
\int^{pi/6}_{0} 1/2 (16-4csc^{2}\thetad\theta)
\int^{pi/6}_{0} (8-2csc^{2}\thetad\theta)
(8\theta+2cot\theta)|^{pi/6}_{0}
(8*pi/6)+2cot(pi/6)-(8*0)-2cot(0)
4pi/3+2√3-0-undefined

And that's my problem..the undefined 2cot.

I think where I went wrong is that sinθ=1/2 at \pi/6 and 5\pi/6 and thus my boundaries of integration should be \pi/6 and 5\pi/6.

Am I on the right track?

Thanks for any feedback/assistance.

Jason

You have the intersection points correct but your integrals aren't. Think of $r$ as a circular windshield wiper. As it moves around the circle, most of the time it is 4 units long as it sweeps over the bottom of the circle. However as it sweeps over the top half, when it hits the straight line, it needs to shorten itself with a variable length to sweep just the region under the line. So part of the time $r=4$ and part of the time $r=2\csc\theta$. So you have to break the integral up into two pieces, using the two different values for $r$. And you want to integrate once around the circle in a positive direction. Does that help?

 

[DCBaelar]

LCKurtz,

Thank for the reply. If I understand correctly, I need to integrate 2cscθ from 5π/6 to π/6 and then subtract the integration of the circle from 5π/6 to π/6. Take that value and subtract from the total area of the circle as found with πr^2?

Thanks,

Jason

 

[LCKurtz]

LCKurtz,

Thank for the reply. If I understand correctly, I need to integrate 2cscθ from 5π/6 to π/6 and then subtract the integration of the circle from 5π/6 to π/6. Take that value and subtract from the total area of the circle as found with πr^2?

Thanks,

Jason

I don't know if you wrote what you meant. On the line segment, $\frac 1 2 r^2$ isn't $2\csc\theta$. And why mess around with all that adding and subtracting areas. You could go once around in the positive direction by starting at $-\frac{7\pi} 6$ and going counterclockwise to $\frac \pi 6$ for the circle part and $\frac \pi 6$ to $\frac{5\pi} 6$ for the other part. Remember you must always integrate in the positive $\theta$ direction.