Question

Evaluate the following integral. Integral from nothing to nothing Subscript 0 Superscript 1 Baseline Integral from nothing to nothing Subscript 0 Superscript pi Baseline Integral from nothing to nothing Subscript 0 Superscript pi divided by 6 Baseline 48 rho sine cubed phi font size decreased by 2 d phi font size decreased by 2 d theta font size decreased by 2 d rho

Answer 1

Answer:

$[16-9\sqrt{3}]\pi$

Step-by-step explanation:

$\int\limits^1_0\int\limits^\pi_0 \int\limits^\frac{\pi}{6}_0 {48\rho Sin^{3}\phi } \, d \phi d\theta d\rho$

splitting it we have

$48 [\int\limits^1_0 {\rho} \, d\rho ] [\int\limits^\pi_0 {1} \, d\theta ] [\int\limits^\frac{\pi}{6} _0 {Sin^{3} \phi } \, d\phi ]$

integrating them separately we have;

$48 [\frac{\rho^2}{2} ]^1_0 [\theta]^\pi_0 [\frac{1}{3} Cos^{3}\phi - Cos\phi + C ]^\frac{\pi}{6}_0$

substituting and evaluating respectively we have;

$48 [\frac{1}{2} ] [\pi] [\frac{-3\sqrt{3} }{8} + \frac{2}{3} ]$

Simplifying we have;

$[16-9\sqrt{3}]\pi$

∴ $\int\limits^1_0\int\limits^\pi_0 \int\limits^\frac{\pi}{6}_0 {48\rho Sin^{3}\phi } \, d \phi d\theta d\rho$ = $[16-9\sqrt{3}]\pi$