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I've difficulties calculating the following sum of integrals:

$$\int_{0}^{\sqrt2}{{\int_{0}^{x}xe^{{(x^2+y^2)}^{3/2}}dy} dx}+\int_{\sqrt2}^{2}{{\int_{0}^{\sqrt{4-x^2}}xe^{{(x^2+y^2)}^{3/2}}dy} dx}$$

My suggestion would be to rewrite this as: $$\int_{0}^{2}{{\int_{0}^{(1/4)π}r^2cos(φ)e^{r^3}dφ} dr}$$

However, the solution manual says the following with respect to $r$: $$0\leq{r}\leq\sqrt2$$ But I can't figure out why. Is there anyone who can give me a hint, or could it be that the solution manual is wrong?

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    You need to reverse the order of integration. – 2017-02-11
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    That is not useful here. – 2017-02-11
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    @Dr.MV You're right. I see that it leads to an integral of the form $\int e^{cy^3}dy$. – 2017-02-11
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    The solution manual is wrong. The domain is a quater circle. By the way, rhe upper limit should be $\pi/4$, not $1/4\pi$. – 2017-02-11
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    @johnwaylandbayles Yes, the integrand won't be a perfect differential . The domain is just a quater circle and polar coordinates work fine here. – 2017-02-11
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    @Dr.MV Thanks for your reply, I see that I forgot the brackets :) – 2017-02-11

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Draw the region of integration - it ends up just being a sector of a circle. The integral is equal to, in polars,

$$\int_0^2 dr \, r^2 \, e^{r^3} \int_0^{\pi/4} d\phi \, \cos{\phi} = \frac{e^8-1}{3 \sqrt{2}}$$

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    You've got the same answer as I do, which indeed implies that the solution manual is wrong. Thanks for your comment. – 2017-02-11