I am trying to convert the following iterated integral from Cartesian to Cylindrical coordinates:
$$\int_{{\,0}}^{{\,\sqrt{3}}}{{\int_{{\,y}}^{{\sqrt {6 – {y^2}} }}{{\int_{{\sqrt \frac{{{x^2} + {y^2}}}{3} }}^{{\sqrt {8 – {x^2} – {y^2}} }}{{{z^3}\,dz}}\,dx}}\,dy}} $$
I tried plotting the bounds of integration, and here's how I converted them:
The boundaries projected along the $xy$-plane is the region $R$ bounded below by $x=y$ and above by $\sqrt{6-y^2}$, intersecting at $(\sqrt{3},\sqrt{3})$
Hence,
$r$ tends from $0$ to $\sqrt{3}$,
$\theta$ tends from $\frac{\pi}{4}$ to $\frac{\pi}{2}$,
$z$ from $\frac{r}{\sqrt{3}}$ to $\sqrt{8-r^2}$.
Now, the converted iterated integral is given by
$$\int_{{\, \frac {\pi }{4}\;}}^{{\,\frac{\pi }{2}\;}}{{\int_{{\,0}}^{{\,\sqrt{3}}}{{\int_{{\,{\frac{r}{\sqrt{3}}}}}^{{\,\sqrt{8-r^2}}}{{z^3{r}\,dz}}\,dr}}\,d\theta }}$$
However, when I try to evaluate both iterated integrals into Wolfram, I don't get the same result: Cartesian result – Cylindrical result
Where could have I gone wrong with my conversion?
Best Answer
Try $0\le\theta\le\pi/4$ and $0\le r\le\sqrt6$. I have no idea where you got your "hence" limits on $r$ and $\theta$.