evaluate integral in spherical coordinates

Question

I thought I knew how to evaluate this, I ended up getting $9\pi/4$ twice. I simply followed a similar example in the book. I can't tell if I'm missing something fundamental or not. Can anyone show me how they get the right answer? thanks.

Answer 1

$$\begin{align}\int_0^\frac\pi6\int_0^\frac\pi2\int_0^3\rho^2\sin\phi\,d\rho\,d\theta\,d\phi &= \frac{\pi}{2}\int_0^\frac\pi6\int_0^3 \rho^2\sin\phi\,d\rho\,d\phi \\ &=\frac{\pi}{2}\left[\frac{\rho^3}{3}\right]_0^3\int_0^\frac\pi6 \sin\phi\,d\phi\\ &=\frac{9\pi}{2}\int_0^\frac\pi6 \sin\phi\,d\phi\\ &=\frac{9\pi}{2}\left[-\cos\phi\right]_0^{\frac\pi6}\\ &=\frac{9\pi}{2}\left(1-\frac{\sqrt{3}}2\right)\\ &=\frac{9\pi}{4}(2-\sqrt{3}) \end{align}$$

Answer 2

\begin{align}\int_0^\frac\pi6\int_0^\frac\pi2\int_0^3\rho^2\sin\phi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\phi&=\left(\int_0^\frac\pi6\sin\phi\,\mathrm d\phi\right)\left(\int_0^\frac\pi21\,\mathrm d\theta\right)\left(\int_0^3\rho^2\,\mathrm d\rho\right)\\&=\left(1-\frac{\sqrt3}2\right)\times\frac\pi2\times9\\&=\frac{9\pi}2\left(1-\frac{\sqrt3}2\right).\end{align}