Answer to Question #108196 in Calculus for Garima Ahlawat

Question

Evaluate triple integration of cos(x^2 + y^2 + z^2)^3/2 dx dy dz , bounded by the sphere x^2 + y^2 + z^2 = 25

Answer 1

ANSWER: $\frac { 2\pi }{ 3 } \cdot \sin { 125\quad }$

EXPLANATION.

Replace the Cartesian coordinate s with spherical. We denote

$D=\left\{ \ \left( x,y,z \right) :\ ^{ }{ x }^{ 2 }{ +y }^{ 2 }+\ { z }^{ 2 }\le 25 \right\}$ , $\Delta =\left\{ (\rho ,\theta ,\varphi ):0\ \le \ \rho \le 5,0\le \varphi \le \pi ,0\le \theta \le 2\pi \right\}$ .

$\iiint _{ D }^{ }{ \cos { { \left( { x }^{ 2 }{ +y }^{ 2 }+\ { z }^{ 2 } \right) }^{ \frac { 3 }{ 2 } \ }{ dxdydz }^{ } } }$$=\iiint _{ \Delta \ }^{ }{ { (\rho \ }^{ 2 }\sin { \varphi } )\cos { { \rho }^{ 3 } } d\rho \ d\varphi d\theta }=$ $=\int _{ 0 }^{ \pi }{ \int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ 5 }{ { (\rho \ }^{ 2 }\sin { \varphi } )\cos { { \rho }^{ 3 } } d\rho d\theta } d\varphi } } =$ $=2\pi \ \left( \int _{ 0 }^{ \pi }{ \sin { \varphi } d\varphi } \right) \cdot \left( \int _{ 0 }^{ 5 }{ { (\rho \ }^{ 2 }\cos { { \rho }^{ 3 } } \ )\ d\rho \ } \right) =$ $=2\pi (-\cos { \pi + } \cos { 0) } \cdot \ \frac { 1 }{ 3 } \left( \sin { 125- } \sin { 0 } \right) =\frac { 2\pi }{ 3 } \cdot \sin { 125\quad }$

Note $\int { \sin { \varphi } d\varphi =-\cos { \varphi } +C,\quad } \int { { (\rho \ }^{ 2 }\cos { { \rho }^{ 3 } } \ )\ d\rho } =\frac { 1 }{ 3 } \int { \cos { { \rho }^{ 3 } } d{ \rho }^{ 3 }\ =\frac { 1 }{ 3 } \sin { { \rho }^{ 3 }+C } } \quad \\ \\ \\ \quad$