Is this integral set up correctly?

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The discussion focuses on calculating the area of the region within the circles defined by the polar equations r=cos(θ) and r=sin(θ) using a double integral. The user proposed the integral 2∫(π/2)^(π/4) ∫(cos(θ))^0 r dr dθ, multiplying by 2 to account for the area being only half of the total. While this approach is valid, it is noted that using the formula A = (1/2) ∫ r² dθ would be simpler, although the assignment specifically required the use of a double integral. The user expresses confidence that their expression will yield a correct numerical result.

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Find the area of the region within both circles;

r=cos(theta) and
r=sin(theta)

using a double integral.

I made the following:

[tex]2\int^{\pi/2}_{\pi/4} \int^{\cos\theta}_0 r dr d\theta[/tex]

I multiplied by 2 because the area I have is only half of the total area to be found. Is this correct or am i doing something stupid?
 
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I'm not sure that it's a particularly good example for using multiple integrals, but you're effectively using Green's Theorem on something that can be readily handled by:
[tex]A = \frac{1}{2} \int r^2 d\theta[/tex]
or, for those who prefer not to deal with polar integrals:
[tex]A=2 \int_0^{\frac{1}{2}} \left(\sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}-x \right)dx[/tex]

I'm fairily sure that your expression will give a numerically correct result.

http://mathworld.wolfram.com/GreensTheorem.html
 
Last edited:
Thanks Nate, I agree that its probably easier to do this with
[tex]1/2 \int r^2 d\theta[/tex] but our assignment was to do it using a double integral. Thanks for the help. I'll keep working on it and see if i get the right answer.
 

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