Double Integral Over a Region: Finding Limits of Integration

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The discussion centers on finding the limits of integration for the double integral \iint\limits_D x{\rm{d}}x{\rm{d}}y, with the region defined by x = \sqrt{2y - y^2} and y = \sqrt{2x - x^2}. The proposed limits are \int_0^1 \int_{1-\sqrt{1-y^2}}^{\sqrt{2y-y^2}} x {\rm{d}}x{\rm{d}}y, with the inner integral focusing on the bounds for x and the outer integral for y. An alternative approach of integrating with respect to y first is mentioned, although it is considered more time-consuming. Additionally, converting the equations to polar coordinates is suggested as a more efficient method for evaluating the integral. The main challenge highlighted is determining the correct bounds for integration.
nuuskur
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Homework Statement


\iint\limits_D x{\rm{d}}x{\rm{d}}y where x = \sqrt{2y - y^2}, y = \sqrt{2x - x^2}

Homework Equations

The Attempt at a Solution


I have figured out the region in question:
jlC2Xbj.png
But how do I get the limits of integration?

Ah, perhaps..
\int_0^1 \int_{1-\sqrt{1-y^2}}^{\sqrt{2y-y^2}} x {\rm{d}}x{\rm{d}}y
In the inner integral I consider what X is bound by and then the 2nd integration would be just from 0 to 1 since that's what y is bound by. Not really worried about the actual calculation itself, but the most challenging bit is figuring out the bounds.
 

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yep. those bounds look good. I get the same ones at least :)

edit: you could also do the integration in the opposite order (i.e. dy first), but I think that way is more time-consuming. and of course, in that method, the limits would look similar, but would have x instead of y.
 
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@nuuskur: Of course, if you were "worried" about evaluating the integral, it would be wise to convert the equations to polar coordinates and set it up and work it that way.
 
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