Rewrite the equation - 5x2+3x+5y2+5y-3z2+4z+12=0 - 5x2+3x+5y2+5y-3z2+4z+12=0 in cylindrical and spherical coordinates. NOTE: write any greek letters using similar standard characters - i. e., for θθ use t, for rhorho use r, for ϕϕ use f, etc.

Cylindrical:

5r^2 (sin (t) ^2 - cos (t) ^2) + r (3cos (t) + 5sin (t)) - 3z^2 + 4z + 12 = 0

Spherical:

5r^2sin (t) ^2 (sin (f) ^2 - cos (f) ^2) - 3r^2 cos (t) ^2 + r sin (t) (3cos (f) + 5sin (f)) + 4r cos (t) + 12 = 0

Step-by-step explanation:

In cylindrical coordinates:

x = r cos (t)

y = r sin (t)

z = z

Let us reorganize the original equation

-5x^2+3x+5y^2+5y-3z^2+4z+12=0

5 (y^2-x^2) + 3x + 5y - 3z^2 + 4z + 12 = 0

Now, we can replace x and y:

5 (r^2 sin (t) ^2 - r^2 cos (t) ^2) + 3rcos (t) + 5r sin (t) - 3z^2 + 4z + 12 = 0

5r^2 (sin (t) ^2 - cos (t) ^2) + r (3cos (t) + 5sin (t)) - 3z^2 + 4z + 12 = 0

In spherical coordinates:

x = r sin (t) cos (f)

y = r sin (t) sin (f)

z = r cos (t)

Let us reorganize the equation:

5 (y^2-x^2) - 3z^2 + 3x + 5y + 4z + 12 = 0

5r^2sin (t) ^2 (sin (f) ^2 - cos (f) ^2) - 3r^2 cos (t) ^2 + r sin (t) (3cos (f) + 5sin (f)) + 4r cos (t) + 12 = 0