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I want to calculate the integral over $$A=\{ (x,y): \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \}$$

When I solve that equation for y, I get:

$$ y\leq b\sqrt{1-\frac{x^2}{a^2}}$$

But how can x be bounded, when I integrate y in this way at first?

Steven33
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    No, you get $-b\sqrt{1-x^2/a^2}\le y\le b\sqrt{1-x^2/a^2}$. And then $1-x^2/a^2\ge 0$, i.e., $x^2\le a^2$ or $-a\le x\le a$. So, the area is $2b\int_{-a}^a\sqrt{1-x^2/a^2},dx$. – amsmath Oct 09 '19 at 18:01
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    https://math.stackexchange.com/questions/548876/find-the-area-of-the-ellipse/1891110 – StubbornAtom Oct 09 '19 at 18:17
  • I see. You get the upper bound for x by looking where x is undefined? – Steven33 Oct 09 '19 at 18:38
  • Nowhere in the Question is it clear that you are trying to find the area, because "calculate the integral" might mean the integral of an arbitrary integrand over the given elliptical region. Please reply if a different interpretation than area was intended. – hardmath Oct 11 '19 at 01:40

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