Area of circle using integration

  • #1

Homework Statement


http://postimage.org/]Screen_Shot_2016_01_09_at_10_13_38_AM.png [/PLAIN] [Broken]
free picture upload


2. The attempt at a solution
I want to go width times delta height. To do this I must describe width in terms of height.

Here they used the Pythagorean theorem which is weird to me because I don't see a nice triangle. Where h meets w the triangle extends outside of the circle. We divide w by 2 I guess to treat it as like a radius?

I guess my question is how do I describe width in terms of height in this context?
 
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Answers and Replies

  • #2
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Homework Statement


Screen_Shot_2016_01_09_at_10_13_38_AM.png [/PLAIN] [Broken]
free picture upload


2. The attempt at a solution
I want to go width times delta height. To do this I must describe width in terms of height.

Here they used the Pythagorean theorem which is weird to me because I don't see a nice triangle.
The triangle isn't shown in the picture. The vertical leg of the triangle goes from the center up to the horizontal strip. The horizontal leg is half the width of the strip. The hypotenuse is the radius of the circle.
PhysicsBoyMan said:
Where h meets w the triangle extends outside of the circle. We divide w by 2 I guess to treat it as like a radius?
No. See above.
PhysicsBoyMan said:
I guess my question is how do I describe width in terms of height in this context?
They already give you w in terms of h in the equation ##w = 2\sqrt{3^2 - h^2}.##
 
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