Surface area of smooth parametric surface

[zzheng]

Sorry I am new to the forum, I don't know how to type in the integrals and stuff.

Homework Statement

Let S be the portion of the surface y = x² where 0 <= z <= X <= 2. Compute the surface area of S.

Homework Equations

r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k
A(S) = integral r_u X r_v dA

The Attempt at a Solution

My first attempt I did: x = ucosv, y = u²cos²v, and z = ucosv where 0<= u <= 2 and 0 <= v <= 2pi
when I was doing r_u X r_v, got a lot of sines and cosines but everything canceled out at the end and became 0.

Then I just decided to use x = x, and y = x² where x is from 0 to 2.
but then r_y turns out to be 0.

I am not sure how I would do this, most of the problems I have done are z as a function of x and y, but this case z is given but I don't know what to do with it.

Could some one please help me with this? Thank you for your help in advance.

 

[LCKurtz]

Unless you are describing a surface of revolution, which you haven't described, I don't see why you would involve trig functions in the parameterization. The surface y = x² is a cylindrical parabola standing vertically on the xy plane. If I understand your description correctly, you are talking about that portion of the cylindrical surface in the first octant under the plane z = x.

If I'm correct, the natural paramaterization would be to use x and z:

\vec R(x,z) = \langle x,x^2,z\rangle

and your domain is a triangle in the xz plane. Try that.