Question

Evaluate the appropriate triple integral

Answer 1

In spherical coordinates, the region R is the set

$R = \left\{ (\rho, \theta, \varphi) \, : \, 0 \le \rho \le 3, \, 0 \le \theta \le 2\pi, \, \dfrac\pi6 \le \varphi \le \pi\right\}$

Then the volume of R is

$\displaystyle \iiint_R dV = \int_0^3 \int_0^{2\pi} \int_{\frac\pi6}^\pi \rho^2 \sin(\varphi) \, d\varphi \, d\theta \, d\rho$

The integrand is free of $\theta$, so we can immediately compute that integral and pull out a factor of 2π :

$\displaystyle \iiint_R dV = 2\pi \int_0^3 \int_{\frac\pi6}^\pi \rho^2 \sin(\varphi) \, d\varphi \, d\rho$

and the remaining double can be factorized as

$\displaystyle \iiint_R dV = 2\pi \left(\int_0^3 \rho^2 \, d\rho\right) \left(\int_{\frac\pi6}^\pi \sin(\varphi) \, d\varphi\right)$

The single integrals are trivial:

$\displaystyle \int_0^3 \rho^2 \, d\rho = \frac13 (3^3 - 0^3) = 3^2 = 9$

$\displaystyle \int_{\frac\pi6}^\pi \sin(\varphi) \, d\varphi = -\cos(\pi) - \left(-\cos\left(\frac\pi6\right)\right) = 1 + \frac{\sqrt3}2$

So, the volume of R is

$\displaystyle \iiint_R dV = 2\pi \cdot 9 \cdot \left(1 + \frac{\sqrt3}2\right) = \boxed{9(2+\sqrt3)\pi}$