If a surface of revolution is obtained by rotating the curve y = g(z) (a lesstha

Question

If a surface of revolution is obtained by rotating the curve y = g(z) (a lessthanorequalto z lessthanorequalto b) about the z-axis, then its equation is x^2 + y^2 = (g(z))^2, and a convenient parameterization of the surface is given by x = g(z)cos theta, y = g) z) sin theta, z = z, 0 lessthanorequalto theta lessthanorequalto 2 pi, a lessthanorequalto z lessthanorequalto. Show that a geodesic on such a surface, given by theta = theta(z), has equation theta(z) = c_1 integral squareroot 1 + (g'(z))^2/g(z) squareroot g(g(z))^2 - c_1^2 dz + c_2

Explanation / Answer

For a Surface of revolution, the geodesic and the meridians are same hence we can find the meridian of the equation given and implies that given equation for geodesic

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