# My professor said that this integral = 0

He said that the integral below equal to zero by an easy application of the divergence theorem.

$$\int_{\Omega }^{}f^{5}(f_{x}+2f_{y})=0$$

where f is some function

Can somebody explain how?

Hello Weilin
What does the integration area omega mean here? Perhaps I can try to solve your question if I know that.
greetings Janm

In this case, omega is just some closed domain lying in R^2

In this case, omega is just some closed domain lying in R^2
So the equation has no poles = singularities?

ah yes, the function is continuous and the derivatives are continuous too. It's basically a very nice domain to work with. no strings attached.

ah yes, the function is continuous and the derivatives are continuous too. It's basically a very nice domain to work with. no strings attached.
Hello Weilin
my function theory was a lot of years ago and my book complex variables is somewhere lost for a long time. At the time I bought it together with a collegue student... The same thing as with comic books. Just the nice ones get lost... I still want to find a intuitive answer to your question. May I do that with a counter question.
Would the integral of f^4*(2f_x+f_y) also be zero?

So has it to do with odd or even powers of f? If so we could reduce it to the integral f*(f_x+2f_y)...

greetings Janm

Yes I think in that case it still would be zero. I think the main thing is the partial derivatives in the parenthesis. I'd imagine you would manipulate that somehow with the divergence theorem to make that parenthesis zero...That is my guess at least.

So what if you put f=x, is it then zero? Not as far as I can tell, if Omega is any domain in R^2.

In this case, omega is just some closed domain lying in R^2

Is it a standard double integral over the region omega?

$$\int_{\Omega}\int f^{5}\left(f_{x}+2f_{y}\right)dA$$

If so, then it is not necessarily 0. Please give more notation, path integral, line integral, double integral, Mdx + Ndy, F dot ds, F dot n ds, etc...

So what if you put f=x, is it then zero? Not as far as I can tell, if Omega is any domain in R^2.

Hello daudaudaudau
if you put f(z)=f(x+iy)=Re(z)=x, then f_x=1 and f_y=0 so the integrand would be x^5(1+2*0)=x^5.
You get the integral of x^5 dA
f(z)= x^5 is a surface bended in one direction and can be made of flat paper.
greetings Janm