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:Cylindrical:5r^2(sin(t)^2 - cos(t)^2) +r(3cos(t) + 5sin(t)) - 3z^2 + 4z + 12 = 0Spherical: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 = 0Step-by-step explanation:In cylindrical coordinates:x = r cos(t)y = r sin(t)z = zLet us reorganize the original equation−5x^2+3x+5y^2+5y−3z^2+4z+12=05 (y^2-x^2) + 3x + 5y - 3z^2 + 4z + 12 = 0Now, 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 = 05r^2(sin(t)^2 - cos(t)^2) +r(3cos(t) + 5sin(t)) - 3z^2 + 4z + 12 = 0In 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 = 05r^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