Parametric Surfaces: Integral of S = x^2 + y^2 + 2z^2 = 10

[Black Orpheus]

I need to take a surface integral where S is
x^2 + y^2 + 2z^2 = 10. I need help with the parametrization of the curve. Letting x=u and y=v makes the problem too complicated. Can you let x=cos(u), y=sin(u) and z=3/sqrt(2)?

 

[0rthodontist]

If z = 3/sqrt(2), then z is a fixed number. You only have one argument, u. So that's going to give you a curve with a fixed z-coordinate when you want a surface.

You know you have an ellipsoid. One way to do this is to transform the ellipsoid into a sphere (a linear transformation--think geometrically), then use the spherical coordinates transformation. So your total transformation could be the composition of the transformation from spherical coordinates and the transformation from a sphere.

 

[HallsofIvy]

I need to take a surface integral where S is
x^2 + y^2 + 2z^2 = 10. I need help with the parametrization of the curve. Letting x=u and y=v makes the problem too complicated. Can you let x=cos(u), y=sin(u) and z=3/sqrt(2)?

No, you can't for the very obvious reason that z is not a constant! It also has only one parameter where a surface integral requires 2. You can set [itex]x= \sqrt{10}cos(u)sin(v)[/itex], [itex]y= \sqrt{10}sin(u)sin(v)[/itex] and then put that into the equation: x²+ y²+ 2z²= 10sin²+ 2z²= 10. Hmm, suppose we let 2z²= 10cos²v? In other words, [itex]z= \sqrt{5}cos v][/itex] Then the equation reduces to
10= 10 which is true. That is, those are parametric equations for the surface. Do you see how this uses spherical coordinates and varies it appropriately?