Convert the point $( - 7,7 \sqrt { 3 } , 14 ) \text { from rectangular coordinates to spherical }$ coordinates.
A) $\left( 14 \sqrt { 2 } , \frac { \pi } { 4 } , \frac { 2 \pi } { 3 } \right)$
B)$\left( 28 \sqrt { 2 } , \frac { 2 \pi } { 3 } , \frac { \pi } { 4 } \right)$
C) $\left( 14 \sqrt { 2 } , \frac { 2 \pi } { 3 } , \frac { \pi } { 4 } \right)$
D) $\left( 28 \sqrt { 2 } , \frac { \pi } { 4 } , \frac { 2 \pi } { 3 } \right)$
E) $\left( 14 \sqrt { 2 } , \frac { \pi } { 3 } , \frac { \pi } { 4 } \right)$

Question

Convert the point $( - 7,7 \sqrt { 3 } , 14 ) \text { from rectangular coordinates to spherical }$ coordinates. 
A) $\left( 14 \sqrt { 2 } , \frac { \pi } { 4 } , \frac { 2 \pi } { 3 } \right)$
B)$\left( 28 \sqrt { 2 } , \frac { 2 \pi } { 3 } , \frac { \pi } { 4 } \right)$ 
C) $\left( 14 \sqrt { 2 } , \frac { 2 \pi } { 3 } , \frac { \pi } { 4 } \right)$
D) $\left( 28 \sqrt { 2 } , \frac { \pi } { 4 } , \frac { 2 \pi } { 3 } \right)$
E) $\left( 14 \sqrt { 2 } , \frac { \pi } { 3 } , \frac { \pi } { 4 } \right)$

Quizplus · Accepted Answer

The spherical coordinates are $(\rho, \theta, \phi)$ where $\rho = \sqrt{x^2 + y^2 + z^2}$, $\theta = \arccos(z/\rho)$, and $\phi = \arctan(y/x)$. Calculating these gives $\rho = 14\sqrt{2}$, $\theta = \frac{2\pi}{3}$, and $\phi = \frac{\pi}{4}$.

Convert the Point (−7,73,14) from rectangular coordinates to spherical ( - 7,7 \sqrt { 3 } , 14 ) \text { from rectangular coordinates to spherical }(−7,73​,14) from rectangular coordinates to spherical

Convert the Point $( - 7,7 \sqrt { 3 } , 14 ) \text { from rectangular coordinates to spherical }$