Question

Cal 3 Problem:

Answer 1

\[\rho=2\csc \phi \rightarrow \rho=\frac{ 2 }{ \sin \phi }\rightarrow \rho \sin \phi=2 \rightarrow \rho^2 \sin^2 \phi=4 \rightarrow \rho^2 - \rho^2 \cos^2 \phi=4 \rightarrow \]
\[\rightarrow x^2+y^2+z^2-z^2=4 \rightarrow x^2+y^2=4\]

Answer 2

Can anyone explain how the x^2, y^2, and z^2 happened?

Answer 3

ro= sqrt(x^2+y^2+z^2)
phi is so called polar angle. Angle that position vector makes with z axis. So
ro^2cos^2 phi =z^2

Answer 4

It is just the transformation from spherical to cartesian coordinates

Answer 5

got it?

Answer 6

\[\Large \rho^2=x^2+y^2+z^2\]

Answer 7

writting it now to make sense of it

Answer 8

ok i understand but how did the \[-\rho^2\cos^2\phi\] come about?