Surface area proof (vector analysis)

[randomcat]

edit: vector* analysis; sorry for the typo

Homework Statement

Given that A = ||r_u||², B = r_u\bulletr_v, C = ||r_v||²
surface area of S is Area(S) = \int^{d}_{c}\int^{b}_{a} = sqrt (AC - B²) dudv

The Dirichlet energy can be thought of as a function as follows

E(S) = 1/2\int^{d}_{c}\int^{b}_{a} = (||r_u||²+||r_v||²) dudv

Show that Area(S) ≤ 1/2 E(S) and that equality holds if
||r_u||² = ||r_v||² and r_u \bullet r_v = 0

Homework Equations

The Attempt at a Solution

Given that ||r_u||² = ||r_v||² and A = ||r_u||² I know that
E(s) = 1/2\int^{d}_{c}\int^{b}_{a} = (2A) dudv
= \int^{d}_{c}\int^{b}_{a} = (A) dudv


Since A = C, and B = 0, then Area(S) = Area(S) = \int^{d}_{c}\int^{b}_{a} = sqrt (A²) dudv

which is just
\int^{d}_{c}\int^{b}_{a} = A dudv

So I showed that they're equal, but how do I justify that Area(S) is less than 1/2 E(S)?

 

[haruspex]

Looks to me that it should not depend on the integration, i.e. If it is going to be true then it should be provable by comparing the integrands.
So investigate (||r_u||²+||r_v||²) ²

 

[randomcat]

Looks to me that it should not depend on the integration, i.e. If it is going to be true then it should be provable by comparing the integrands.
So investigate (||r_u||²+||r_v||²) ²

Hmm...I'm not sure where that leads me in the proof, could you please elaborate?

 

[haruspex]

(Just noticed that in every one of your integrals the integral sign is separated from the integrand by an equals sign. I'm assumng that's a typo. If not, there's somethng in your notation I'm not familiar with. Also, you seem to have too many 1/2s floating around. Either there's a 1/2 in the definition of E(S) or there's a 1/2 in the inequality to be proved, not both.)

Compare the squares of the integrands, i.e. compare (||r_u||²+||r_v||²) ²/4 with AC-B². Can you see that one is guaranteed at least as large as the other?