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26 November, 21:14

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.

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  1. 26 November, 23:24
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    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
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