Question

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.

Answer 1

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

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