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I'm trying to change the order of integration of

$$\int^{2a}_{0}\left( \int^{\sqrt{2ax}}_{\sqrt{2ax-x^{2}}} f(x,y) dy \right) dx $$

but I'm not sure how to express the lower boundary for y in terms of x.

Thank you.

SOLUTION ~ So here is my final solution: $$\left( \int^{a}_{0} \int^{a-\sqrt{a^{2}-y^{2}}}_{\frac{y^{2}}{2a}}+\int^{a}_{0} \int^{2a}_{a+\sqrt{a^{2}-y^{2}}}+\int^{2a}_{a} \int^{2a}_{\frac{y^{2}}{2a}}\right) f(x,y) dx dy$$

I hope it's correct and thanks for the help.

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  • $\begingroup$ Hint: draw the integration limits and you will see. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 23 '16 at 17:59
  • $\begingroup$ I did, but still don't know how to express the curve in terms of x=f(y) $\endgroup$ – srmico Jan 23 '16 at 18:00
  • $\begingroup$ It is difficult to change the order of integration and stay with one integral. If you draw the region you will see that it is easier to split it into three regions first: upper, lower-left, and lower-right. $\endgroup$ – Rory Daulton Jan 23 '16 at 18:14
  • $\begingroup$ @RoryDaulton yes I did split it in three integrals, thank you. $\endgroup$ – srmico Jan 23 '16 at 18:38
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Hint: $$y = \sqrt{2ax}\implies x =y^2/2a$$ $$y = \sqrt{2ax-x^2}\implies y^2 = 2ax-x^2\implies \cdots$$ (second degree equation)

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  • $\begingroup$ so x=a +/- sqrt(a^2-y^2)? and since y=0 -> x=0 no matter what a, I just take a-sqrt(a^2-y^2)? Alright thank you so much :D $\endgroup$ – srmico Jan 23 '16 at 18:20
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    $\begingroup$ @srmico, you will need both branches. Do the drawing $\endgroup$ – Martín-Blas Pérez Pinilla Jan 23 '16 at 18:28
  • $\begingroup$ Yep, I realized that. I can solve the problem now. Thanks. $\endgroup$ – srmico Jan 23 '16 at 18:37

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