Proof of Integral Relation for Scalar Functions with Gradient

Benny
Messages
577
Reaction score
0
Hi, I'm stuck on the following question.

Q. Let f(x,y,z) and g(x,y,z) be C^2 scalar functions. Let D be an elementary region in space and \partial D be the closed surface that bounds D. Prove that:

<br /> \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_D^{} {\nabla f \cdot \nabla g} } } dV = \int\limits_{}^{} {\int\limits_{\partial D}^{} {f\nabla g \cdot d\mathop S\limits^ \to } } - \int\limits_{}^{} {\int\limits_{}^{} {\int\limits_D^{} {f\nabla ^2 } gdV} } <br />

I've had practice doing some identity problems involving grad, div and curl. In those questions I've just started with the definition and have usually managed to get to the required result. However, I don't know where to start with this one. I can't think of any theorems which I could use to get things going. Can someone please help me get started? Thanks.
 
Last edited:
Physics news on Phys.org
This is a higher dimensional analog of integration by parts, which is the integral form of the differential product rule. See if a product rule helps you here.
 
Thanks for the hint StatusX, I'll see what I can come up with.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top