Double Integral Over a Region: Finding Limits of Integration

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SUMMARY

The discussion focuses on determining the limits of integration for the double integral \(\iint\limits_D x \, dx \, dy\) over the region defined by the curves \(x = \sqrt{2y - y^2}\) and \(y = \sqrt{2x - x^2}\). The participants concluded that the limits of integration can be expressed as \(\int_0^1 \int_{1-\sqrt{1-y^2}}^{\sqrt{2y-y^2}} x \, dx \, dy\). An alternative method of integration, switching the order to \(dy\) first, was discussed but deemed more time-consuming. The suggestion to convert to polar coordinates for easier evaluation was also mentioned.

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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.
 
Last edited:
@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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