Surface area of a parametric surface

[electroguy02]

1. Homework Statement [/b]

For the surface with parametric equations r(s,t) = <st, s+t, s-t>, find the equation of the tangent plane at (2,3,1).

Find the surface area under the restriction s^2 + t^2 <= 1.

2. Homework Equations [/b]



3. The Attempt at a Solution [/b]

I already found that the equation of the plane is -2x+3y-z=4. I just don't know what to do with that parametrization to find the surface area. I tried plugging in the values of r(s,t) into x, y, and z, then integrating after converting s and t (I tried s = cos(theta) and t = sin(theta)) to polar, but that didn't work.

Any ideas on how to get started?

 

[mighty2000]

Hi there,

To find the surface area under the restriction s^2 + t^2 <= 1, you can use the formula for surface area in terms of parametric equations:

S = ∫∫√(r_s^2 + r_t^2 + 1) ds dt

where r_s and r_t are the partial derivatives of r with respect to s and t, respectively.

In this case, we have r_s = <t, 1, 1> and r_t = <s, 1, -1>. So the integral becomes:

S = ∫∫√(s^2 + t^2 + 3) ds dt

To integrate this, you can use the substitution s = cos(theta) and t = sin(theta), which will give you:

S = ∫∫√(cos^2(theta) + sin^2(theta) + 3) cos(theta) dtheta dtheta

= ∫∫√(4) cos(theta) dtheta dtheta

= 4∫∫cos(theta) dtheta dtheta

= 4∫sin(theta) dtheta dtheta

= 4∫-cos(theta) dtheta

= -4cos(theta) + C

= -4cos(arccos(s)) + C

= -4s + C

Now, to evaluate this integral, we need to find the limits of integration. Since we are restricting s^2 + t^2 <= 1, this means that s and t can take on values between -1 and 1. So our limits of integration for s and t are both -1 and 1.

Therefore, the surface area under the restriction s^2 + t^2 <= 1 is:

S = -4s + C = -4(1) + 4(-1) = -8 units^2

I hope this helps! Let me know if you have any further questions. Good luck!