using polar coordinates to find area

Erin0702

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2a. Use polar coordinates to find the area of the region bounded by the two circles x²+y²=1 and x²+y²=4

2b.Then find the same area by elementary geometry to check your integration.



When converting to polar coordinates
r=x+y
so r²=x²+y²
Then I would have r²=4 and r²=1 which would give me r=±2 and r=±1.

From here I'm not sure where to go with it. Any help is greatly appreciated!
 
Erin0702 said:
When converting to polar coordinates, r=x+y....
No. The hypotenuse is not the sum of the two legs.

Erin0702 said:
...so r²=x²+y²
This statement is correct, but does not follow from the previous (incorrect) statement, since (x + y)<sup>2</sup> = x<sup>2</sup> + 2xy + y<sup>2</sup>, which is not the same as x<sup>2</sup> + y<sup>2</sup> = r<sup>2</sup>.

Just use the radii, and find the area, integrating in polar coordinates, between r<sup>2</sup> = 4 and r<sup>2</sup> = 1.

Eliz.
 
My book says that x²+y²=r² and since I'm given the equations x²+y²=4 and x²+y²=1 I thought I could make them r²=±2 and r²=±1?
 
Erin0702 said:
My book says that x²+y²=r² and since I'm given the equations x²+y²=4 and x²+y²=1 I thought I could make them r²=±2 and r²=±1?

It is critical to review algebra before studying calculus. Please see your teacher for extra help. The specific problems you are having with this question are not calculus related.
 
My problem isn't algebra...I'm converting the cartesian coordinates to polar coordinates. There is no algebra involved in that it simply says that x^2+y^2=r^2 everytime. I just made the mistake of also saying that r=x+y. All I needed to know was if I was going about solving this problem in the right way, I wasn't trying to perform any algebraic steps.
 
Erin0702 said:
My problem isn't algebra.
According to what you have posted, you have somehow come to the conclusion that radii can be negative. This is a non-calculus difficulty that needs to be ironed out before proceeding.

Please take the tutor's advice to heart; it was meant to help you.

Thank you.

Eliz.
 
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