Calculus - finite integration of $e^{y^3}$ in double integration

Question

i have this problem that bugs me for 3 hours now. I searched the internet and did not find a solution to this specific problem which was asked in our final:

$$\int_0^3 \;\int_{\sqrt{x/3}}^r e^{y^3}\;dy\;dx$$

i guess this is solved by polar coordinates and jacobien, but i cant seem to find an asnswer. Thanks in advance!

Answer 1

Hint

Maybe the result:

$$\int \exp{y^p} \, \mathrm{d} x = \frac{ \Gamma \left(\frac{1}{p},-y^p\right)}{p} = -\frac{y E_{\frac{p-1}{p}}\left(-y^p\right)}{p}, \quad p \neq 0,$$ where $E_i$ is the exponential integral and $\Gamma$ is the incomplete gamma function; is of any help.

Answer 2

Switch the order of integration.

The computation is a bit messy for a general $r$, but, by way of illustration, if we take $r=1$ we have a simple computation: $$\int_0^3\int_{\sqrt{x/3}}^1e^{y^3}dy\,dx=\int_0^1\int_0^{3y^2}e^{y^3}dx\,dy=\int_0^13y^2e^{y^3}dy=e^{y^3}\Big|_0^1=e-1.$$ Now, try to generalize the computation to any $r$. [Hint: You will have to split the integral into two parts.]