Area of Circle Cut by Line: Find Solution Here

[springo]

Homework Statement

Find the area of the two regions in which y = mx divides x²+y²-2ax < 0.

Homework Equations

The Attempt at a Solution

I was thinking of the following:
\int_{0}^{2a}\int_{\sqrt{2ax-x^2}}^{mx}dy\:dxThanks for your help!

 

[Mark44]

This should work, except that your lower limit of integration on the inner integral should be -\sqrt{2ax - x^2}, since for a given x value, your y values range from the lower half of the circle up to the line.

This integral will give you the area within the circle up to the line. The remaining area is the area of the circle minus the area you found from the integral.

 

[gabbagabbahey]

This integral will give you the area within the circle up to the line.

I think there is a small problem with this idea: Depending on the value of m , the line may (only if m=0) or may not be the upper bound for the entire lower region. If the line y=mx cuts through the disk exactly in half, then your integral will work. However, if it doesn't y=mx will only be the upper bound until the second intersection point (x=\frac{2a}{1+m^2} ); after that, the upper bound will be y=\sqrt{2ax-x^2}.

 

[springo]

Thanks for the correction Mark.

Gabbagabbahey, I'm not entirely sure what you mean. Is it that that the integral only adds area if the line is inside the circle? That would imply that for negative values of m, we could get the area just with this integral, and for positive values of m we could get the area by using \sqrt{2ax-x^2} as upper bound and mx as lower bound, wouldn't it?

 

[Mark44]

I think there is a small problem with this idea: Depending on the value of m , the line may (only if m=0) or may not be the upper bound for the entire lower region. If the line y=mx cuts through the disk exactly in half, then your integral will work. However, if it doesn't y=mx will only be the upper bound until the second intersection point (x=\frac{2a}{1+m^2} ); after that, the upper bound will be y=\sqrt{2ax-x^2}.

You're right. I even have a drawing, but didn't look closely enough at it.

 

[gabbagabbahey]

Thanks for the correction Mark.

Gabbagabbahey, I'm not entirely sure what you mean. Is it that that the integral only adds area if the line is inside the circle? That would imply that for negative values of m, we could get the area just with this integral, and for positive values of m we could get the area by using \sqrt{2ax-x^2} as upper bound and mx as lower bound, wouldn't it?

I'll attach a couple of quick plots to show you what I mean. In the first plot (a=2 and m=2 ), the integral calculates the area of the entire shaded region, not just the area inside the circle. In the second plot (a=2 and m=-2 ) the integral gives the area of the two lightly shaded regions.

http://img265.imageshack.us/img265/1762/circ1.th.jpg

http://img231.imageshack.us/img231/6420/circ2.th.jpg

 

[springo]

OK I see now.
So what can I do then?

 

[gabbagabbahey]

Break it up into two integrals, between x=0 and the intersection point your upper and lower y-bounds are the line and the lower part of the circle. Between the intersection point and x=2a, your upper and lower bounds are the upper and lower halves of the circle.

Alternatively, you could switch to polar coordinates and avoid this problem altogether.

 

[Mark44]

BTW, you don't need an iterated integral to do this problem.

Here's what I would do.

If m > 0, find the point of intersection of the line with the circle, as gabbagabbahey already showed. The inner integral should have the y value range from the line up to the circle, and the outer integral should range from x = 0 to the x value at the other point of intersection. That gives you the area of the upper (and smaller) piece. The area of the other piece is the area of the circle minus what you just found.

If m < 0, change the sign of m and proceed as above. The circle is symmetric with respect to the x-axis, so it doesn't matter if the line is, for example, y = 2x or y = -2x. With either line, the small segments have the same area, and the large segments have the same area.

 

[springo]

Thanks, I solved it by doing it with a regular integral, but since this is in my list of exercizes about double integrals, I think I have to find a way to do it using those.

I was trying the polar method because it seems the "cleanest" way to do it since the integrals are less complicated.
So, I got:
θ = arctan(m) ...the line
r² - 2·a·r·cos(θ) < 0 ...the circle

So...
\int_{\arctan m}^{\frac{\pi}{2}}\int_{0}^{2a\cos \theta}r\:dr\:d\theta

And then just integrate which is pretty easy here...
Is that OK?

 

[gabbagabbahey]

I was trying the polar method because it seems the "cleanest" way to do it since the integrals are less complicated.
So, I got:
θ = arctan(m) ...the line
r² - 2·a·r·cos(θ) < 0 ...the circle

So...
\int_{\arctan m}^{\frac{\pi}{2}}\int_{0}^{2a\cos \theta}r\:dr\:d\theta

And then just integrate which is pretty easy here...
Is that OK?

Looks good to me!

 

[springo]

Ok then, thanks for your help guys!