# Deduction with Stoke's Theorem

#### TheFirstOrder

Hi guys! I hope someone here will be able to help me with this question:

f(r) is a scalar field, then use stoke's theorem (∫∫c curlF.dS=∫cF.dr) to deduce that:

∫∫s grad(f) x dS = -∫c f dr

I am stuck as to how to do this. I have tried subbing in the vector calculus identity curl(fu)=fcurl(u)+(gradf)xu (where u is a constant). Thus

∫∫s grad(f) x dS = ∫∫s curl(fu) - fcurl(u) (where dS is u)

but I am not sure whether this is right (and i then get stuck at this point too). Please help!

#### HallsofIvy

MHF Helper
If u is a constant vector then curl u= 0. But then you say "where dS= u" which makes no sense- dS isn't constant except on planes.

#### TheFirstOrder

If u is a constant vector then curl u= 0. But then you say "where dS= u" which makes no sense- dS isn't constant except on planes.
Yes, I figured out as much. But what would be the first step in the right direction? I am really not sure of what to do.

#### TheFirstOrder

Sorry, I mean to say: what other vector calculus identity can I could use to solve this problem?

#### HallsofIvy

MHF Helper
Sorry, I mean to say: what other vector calculus identity can I could use to solve this problem?
I notice that you have started a second thread for this question- you really shouldn't do that.

#### TheFirstOrder

I notice that you have started a second thread for this question- you really shouldn't do that.
Sorry, I don't quite understand. This question got moved from differential equations because I placed it there by accident, but I began only one thread for this question.

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#### HallsofIvy

MHF Helper
My apology! I didn't realize that two different people were asking exactly the same question! Perhaps you and silverflow are taking the same course?

#### TheFirstOrder

My apology! I didn't realize that two different people were asking exactly the same question! Perhaps you and silverflow are taking the same course?

haha. we must be!