# Surface area proof (vector analysis)

1. Mar 15, 2013

### randomcat

edit: vector* analysis; sorry for the typo
1. The problem statement, all variables and given/known data
Given that A = ||ru||2, B = ru$\bullet$rv, C = ||rv||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}$ = (||ru||2+||rv||2) dudv

Show that Area(S) ≤ 1/2 E(S) and that equality holds if
||ru||2 = ||rv||2 and ru $\bullet$ rv = 0
2. Relevant equations

3. The attempt at a solution
Given that ||ru||2 = ||rv||2 and A = ||ru||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)?

Last edited by a moderator: Mar 16, 2013
2. Mar 15, 2013

### 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 (||ru||2+||rv||2) 2

3. Mar 15, 2013

### randomcat

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

4. Mar 16, 2013

### 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 (||ru||2+||rv||2) 2/4 with AC-B2. Can you see that one is guaranteed at least as large as the other?