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Alright, so I'm stuck on what I think should be a simple double integral. It is $\int_0^1\int_{\sqrt x}^1e^{y^3} \, dy \, dx$. This is just the volume between the surface $z=e^{y^3}$ and the area bounded by $y=\sqrt x$ and $y=1$, $0\le x \le 1$.

I could see pretty quickly that I can't integrate this, so I tried to switch the $dy$ with $dx$ and change limits. When I do that, though, I end up with $\int_0^1[y^2e^{y^3}] \, dy$, which I still can't integrate.

So I try by converting to polar coordinates, and I get that $0\le\theta\le\frac{\pi}{2}$ and $\frac{2\cos\theta}{1-\cos 2\theta}\le r \le 1-\sin\theta$. That radius will be a major problem when I go to integrate, so I know I went wrong somewhere.

Does someone see what I did wrong?

Thanks in advance!

Harish Chandra Rajpoot
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PoGaMi
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    $\int_0^1 y^2e^{y^3} \ dy$ is easy to integrate, as $\frac{d \ }{dy}y^3 = 3y^2$ – Simon S Jul 31 '15 at 04:27
  • Oh wow, can't believe I missed that. Working too hard. Thanks! Can you post as an answer so that I can mark the question closed? – PoGaMi Jul 31 '15 at 04:36

1 Answers1

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Change of order of integration we have $$\int_{0}^{1}\int_{\sqrt{x}}^{1}e^{y^3}dydx$$

By changing order of integration, we get area bounded by $x=y^2$, $x=0$, $0\leq y\leq 1$ $$=\int_{0}^{1}\int_{0}^{y^2}e^{y^3}dxdy$$ $$=\int_{0}^{1}\left(\int_{0}^{y^2}e^{y^3}dx\right)dy$$ $$=\int_{0}^{1}e^{y^3}[x]_{0}^{y^2}dy$$ $$=\int_{0}^{1}e^{y^3}\left(y^2-0\right)dy$$ $$=\int_{0}^{1}y^2e^{y^3}dy$$ Let $y^3=t\implies 3y^2 dy=dt$ $$=\int_{0}^{1}e^{t}\frac{dt}{3}$$ $$=\frac{1}{3}\int_{0}^{1}e^{t}dt$$ $$=\frac{1}{3}[e^t]_{0}^{1}$$ $$=\frac{1}{3}(e^{1}-e^{0})$$$$=\color{blue}{\frac{1}{3}(e-1)}$$

Harish Chandra Rajpoot
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