Need help evaluating double integrals.

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The discussion focuses on evaluating the double integral of the function sqrt(x^2 + y^2) over a specified region defined by the limits from 0 to 2a for y and from -sqrt(2ay - y^2) to 0 for x. The participants suggest converting to polar coordinates to simplify the integration process, leading to the expression r^2 dr d(theta). A key point raised is the importance of correctly identifying the order of integration and understanding the geometric nature of the integration region, particularly the graph of x = -sqrt(2ay - y^2).

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The question states, If a is a positive number, what is the value of the following double integral?

integral from 0 to 2a [ integral from - sqrt(2ay - y^2) to 0 of sqrt(x^2 + y^2) dy dx.My first thought is that we should change to polar coordinates so I do not have to do a trigonometric substitution.

So then it would sqrt(r^2) r dr d(theta) which would simplify to r^2 dr d(theta).

What i am confused with is now how to change my limits of integration to polar? Can someone help please.
 
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I think you have the order of integration switched. Your inner integral is shown as being done with respect to y. I believe the inner integral should be done with respect to x, and that the limits of integration are from x = -sqrt(2ay - y^2) to x = 0. The outer integral is with respect to y, I believe, and the limits are x = 0 to x = 2a.

Possibly the hardest thing about double and triple integrals is understanding the nature of the region over which integration is being done. If you don't understand that, you don't have any chance of being able to change from Cartesian (rectangular) to polar or vice-versa.

What does the region look like? In particular, what does the graph of x = -sqrt(2ay - y^2) look like?
 

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