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I want to calculate the following integral: $$ \int_{-1}^{1}\int_0^2 \sqrt{\left|y-x^2\right|} \,dx \,dy. $$ I tried to go first with $y$ which seems the easier of the two, but then the integral with respect to $x$ becomes quite complex.

On the other hand, starting with $x$ it is cumbersome and further the integral with respect to $y$ is more complex than the first method...

Is there a simple way to do it??

Adrian Keister
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alex vidal
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1 Answers1

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I don't know your method, so it is not possible for me to compare my method and yours' method and decide which is simpler.

Given integral: $$I = \int^{1}_{-1} \int^2_0 \sqrt{|{y-x^2}|} dx dy $$ Let $ f(x,y) = \sqrt{|y-x^2|} $

Here, you have to careful for the values inside $(|.|)$. It should be obvious that for $y \in (-1,0) $ ,$$f(x,y) = \sqrt{| -|y| -x^2 |} = \sqrt{| |y| + x^2 |} $$

And for $ y \in (0,1)$, there are two cases:

  • Case 1: If $ x < \sqrt{y}$, for that $f(x,y) = \sqrt{y-x^2} $
  • Case 2: If $ x \ge \sqrt{y}$, for that $f(x,y) = \sqrt{x^2 -y} $

So,$$I = \int^0_{-1} \int^2_0 \sqrt{ |y| + x^2} + \int^1_0 \int^{\sqrt{y}}_0 \sqrt{y-x^2} dx dy + \int^1_0 \int_{\sqrt{y}}^2 \sqrt{x^2-y} dx dy$$ As the first integral is insenstive to the values of variable $y$, integral $I$ can be written as $$ = \int^1_{0} \int^2_0 \sqrt{ y + x^2} + \int^1_0 \int^{\sqrt{y}}_0 \sqrt{y-x^2} dx dy + \int^1_0 \int_{\sqrt{y}}^2 \sqrt{x^2-y} dx dy$$

Now, all the integrals are of standard form, they can be solved easily, and I leave you that as an exercise.

xrfxlp
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