Surface Area of Parameterized Region

[Destroxia]

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?

 

[Nathanael]

Everything you did looks good to me, but you stopped just one step short of an answer.