Answer to Question #261988 in Electrical Engineering for Billa

Question

convert the following equation written in cartesian coordinate in to an equation of cylindrical coordinates x^3-2x^2-6z=4-2y^2

Answer 1

$x^3 -2x^2+2y^2 -6z-4=0$

Put x = r cos (A)

and, y = r sin (A)

and z =z, we get

$(r cos A)^3 - 2 (r cos A)^2 + 2 (r sin A)^2 - 6z -4=0$

$r^3 cos^3A - 2r^2 cos^2A + 2r^2 sin^2A - 6z-4=0$

$r^3cos^3A - 2r^2 (cos^2A-sin^2A) - 6z-4=0$

$r^3cos^3A -2 r^2cos2A - 6z-4=0$

Final Cylindrical equation:

$r^3cos^3A - 2r^2cos2A - 6z-4=0$