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

`[16-9√(3)]\pi`

Explanation:

`\int\limits^1_0\int\limits^\pi_0 \int\limits^(\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^(\pi)/(6) _0 {Sin^(3) \phi } \, d\phi ]`

integrating them separately we have;

`48 [(\rho^2)/(2) ]^1_0 [\theta]^\pi_0 [(1)/(3) Cos^(3)\phi - Cos\phi + C ]^(\pi)/(6)_0`

substituting and evaluating respectively we have;

`48 [(1)/(2) ] [\pi] [(-3√(3) )/(8) + (2)/(3) ]`

Simplifying we have;

`[16-9√(3)]\pi`

∴
`\int\limits^1_0\int\limits^\pi_0 \int\limits^(\pi)/(6)_0 {48\rho Sin^(3)\phi } \, d \phi d\theta d\rho` =
`[16-9√(3)]\pi`