- #1
Destroxia
- 204
- 7
Homework Statement
Find the surface area S of the portion of the hyperbolic paraboloid:
## r(u,v) = \langle (u+v),(u-v),uv \rangle ##
for which:
## u^2 + v^2 <= 225 ##
Homework Equations
(Surface Area for Parameterized Region:)
##\int \int ||\frac {\partial r} {\partial u} \times \frac {\partial r} {\partial v}||dA## Where the double integral is over the region R.
The Attempt at a Solution
Okay, so first, I found the partial derivatives of u and v with respect to the parameterized function r(u,v):
## \frac {\partial r} {\partial u} = \langle 1,1,v \rangle , \frac {\partial r} {\partial v} = \langle 1,-1,u \rangle ##
After that, I found the cross product:
## \langle (u+v), (v-u), -2 \rangle ##
Then the norm:
## \sqrt{2(u^2+v^2+2)}##
It's at this point that I realized this integrand was pretty rough, and should probably switch to polar:
##\sqrt{2r^2+4}##
Since I got all that set, I need to find some limits, and also convert them to polar coordinates. The limits for u I had as [-3,3], and the limits for v were [##-\sqrt{225-u^2},\sqrt{225-u^2}##]
So, the new limits for the polar system I received for ##\theta## was [0,2\pi], and for r I got [0, 15], due to the circle being at radius 15.
Now I set up the integral and evaluate.
##\int_0^{2\pi} \int_0^15 \sqrt{2r^2+4} r dr d\theta##
Using U substitution I get an integral of this form (I plan on plugging the u back in, so I leave the limits the same):
##\int_0^{2\pi} \int_0^15 (\frac 1 4) \sqrt{u} du d\theta##
##\int_0^{2\pi} (\frac 1 4) [ (\frac 2 3) (2r^2+4)^{\frac 3 2}|_0^{15}]##
The integral just gets super sloppy here, and I'm not sure if I've even set it up correctly, or if you can even convert to polar from a parameterized function?