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# Area Between Polar Curves. Watch

1. I Just have a simple doubt really.

Consider the polar curves r=sinx and r=sin(2x)

This is basically a circle which intersects a four leaf clover two times when you graph it out.

If one were to try and find the intersection points algebraically, you would get pi/3 and 5pi/3

However when you draw it out, the intersection is actually in the second quadrant since r is negative and reflects to the opposite quadrant.

My question now concerns areas regarding these curves. If I wanted to find the area enclosed by the two curves in the second quadrant, how would one go about that?

I'm confused here mainly about the limits of integration. How do you know which angle to which angle to integrate to when looking at the diagram?

If I wanted to find the area from the 0pi x axis till they intersect in quad 2 it would be roughly 2pi/3 degrees but thats 5pi/3 for one of the curves, so how does this work?
2. Think about rays from the origin and consider whether or not, for a particular direction, they pass through the region in question.

For a ray will pass through the region and intersect first.

For a ray will pass through the region and intersect first.

Edit: BTW your 5pi/3 is an error.

The intersections occur when theta = pi/3 and 2pi/3
3. (Original post by BabyMaths)
Think about rays from the origin and consider whether or not, for a particular direction, they pass through the region in question.

For a ray will pass through the region and intersect first.

For a ray will pass through the region and intersect first.

Edit: BTW your 5pi/3 is an error.

The intersections occur when theta = pi/3 and 2pi/3
Actually just using algebra the intersection is at 5pi/3 and that's a correct intersection point, its just that r is negative when your graphing that and it reflects into the second quadrant. So when your calculating the areas in question and in the graph it looks like you want it from pi/2 to pi, do you just input that or do you have to use the real angle values, in this case 1.5pi to 2pi if I'm not mistaken.

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Updated: April 2, 2013
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