Gauss Divergence Theorem - Silly doubt - Almost solved

come2ershad
Messages
16
Reaction score
0

Homework Statement



The problem statement has been attached with this post.

Homework Equations



I considered u = ux i + uy j and unit normal n = nx i + ny j.


The Attempt at a Solution



I used gauss' divergence theorem. Then it came as integral [(dux/dx) d(omega)] + integral [(duy/dy) d(omega)] = integral [(ux nx d(gamma)] + integral [(uy ny d(gamma)]

My question is can I separate the x and y components and write as separate equations as given in the problem? Is that right?
 

Attachments

  • db.jpg
    db.jpg
    23.2 KB · Views: 470
Physics news on Phys.org
anybody?
 
You can, although it is not trivial to prove it. Break up the boundary so that the components of the vector can be written as functions of x on each piece and use x itself as parameter. That will give the first equation.

Then break up the boundary so the components of the vector can be written as functions of y on each piece and use y itself as parameter. That will give the second equation.

That partitioning is used in the proof of the theorem. You might want to look at the proof in any Calculus text.
 
Thanks for the reply.
 
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