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I'm trying to solve this double integral of

f(x,y) = $\frac{1}{\sqrt{1+y^3}}$ on the domain :

  • $0\le$ x $\le$ 1

  • $\sqrt{x}$ $\le$ y $\le$ 1

$$\int_0^1\int_\sqrt{x}^1{\frac{1}{\sqrt{1+y^3}}dxdy}$$

This is where I am stuck, how do I calculate this integral? I couldn't find a good substitution. I have tried transforming to polar coordinates but also to no avail as it seems to make matters worse.

$$\int_0^1(\int_\sqrt{x}^1{\frac{1}{\sqrt{1+y^3}}dy)dx}$$

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  • $\begingroup$ Did you try to integrate with respect to $x$ first? $\endgroup$ – mickep Dec 7 '17 at 13:30
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$$( 0 \le x \le1~~~and~~\sqrt{x}\le y \le1)\Longleftrightarrow ( 0 \le y\le1~~~and~~0\le x\le y ^2 ) $$

Hence By Fubini,

$$\int_0^1\left(\int_\sqrt{x}^1{\frac{1}{\sqrt{1+y^3}}dy}\right)dx=\int_0^1\left(\int_0^{y^2}{\frac{1}{\sqrt{1+y^3}}dx}\right)dy \\= \int_0^1\frac{y^2}{\sqrt{1+y^3}}dx\\=\int_0^1\frac{2}{3}\left(\sqrt{1+y^3}\right)'dx =\color{red}{\frac{2}{3}(\sqrt{2}-1)}$$

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  • $\begingroup$ Thank you very much! This did solve my problem. I did not understand you could swap those, I just started learning about double integrals. This helps a lot. $\endgroup$ – Gert W. Dec 7 '17 at 13:52
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Hint: we have

$\int_0^1(\int_\sqrt{x}^1{\frac{1}{\sqrt{1+y^3}}dy)dx}=\int_0^1(\int_0^{y^2}{\frac{1}{\sqrt{1+y^3}}dx)dy}$.

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