Hello Meryem,

your expressions are indeed somewhat complicated, so I choose a much simpler example which may, at least I hope so, give you a hint.

First of all you should check the "mixed derivatives" of your gradient, that means check the "integrability conditions". That will show if your function f exists.

Now for the example.

Define

grad[f_] := {D[f, x], D[f, y], D[f, z]}

Let's say you have a gradient

gradf = {(2 x)/z, (2 y)/z, -((x^2 + y^2)/z^2)};

Check the mixed derivatives

D[gradf[[1]], y] == D[gradf[[2]], x]
D[gradf[[1]], z] == D[gradf[[3]], x]
D[gradf[[2]], z] == D[gradf[[3]], y]

Each answer is True, so there is indeed a function giving the gradient in question.

Now chose a straight line from { x0, y0, z0 } to { x1, y1, z1 } and the derivative of the line element to calculate

Integrate [ gradf . dr ]

along this line. Set

rule = {x -> x0 (1 - t) + t x1, y -> y0 (1 - t) + t y1, z -> z0 (1 - t) + t z1};
xp = D[{x, y, z} /. rule, t];
df = (gradf /. rule).xp // FullSimplify;

Switch off the output of conditions in Integrate and do the integration, change the variables back to x, y, z and get a function yielding the given gradient

SetOptions[Integrate, GenerateConditions -> False];
ff = Integrate[df, {t, 0, 1}] /. {x1 -> x, y1 -> y, z1 -> z} // FullSimplify
grad[ff]

Of course you may as well integrate along the edges of the cuboid:

fff = Integrate[(gradf[[1]] /. {x -> u, y -> y0, z -> z0}), {u, x0, x1}] +
  Integrate[(gradf[[2]] /. {x -> x1, y -> u, z -> z0}), {u, y0, y1}] +
  Integrate[(gradf[[3]] /. {x -> x1, y -> y1, z -> u}), {u, z0, z1}]
fff = (fff // FullSimplify) /. {x1 -> x, y1 -> y, z1 -> z}
grad[fff]
gradf

I hope you got an idea how to calculate your function f.

In general the procedure may be written as

grad[f_] := {D[f, x], D[f, y], D[f, z]}

t1 = Integrate[a[u, y0, z0], {u, x0, x}];
t2 = Integrate[b[x, u, z0], {u, y0, y}];
t3 = Integrate[c[x, y, u], {u, z0, z}];

(*Integrability conditions *)
rr = {
   Derivative[1, 0, 0][b][xx__] -> Derivative[0, 1, 0][a][xx],
   Derivative[1, 0, 0][c][xx__] -> Derivative[0, 0, 1][a][xx],
   Derivative[0, 1, 0][c][xx__] -> Derivative[0, 0, 1][b][xx]
   };

F = t1 + t2 + t3;
grad[F]
grad[F] /. rr

Hello Hans. I could not deal with this question for a long time due to my health problems. Yesterday I looked at the question again and unfortunately there is no function that gives this gradient as you said.

Thank you so much. I will try to solve the problem using this solution method.

This remains confusing. On one side there is what looks like a surface of the form z=f(x,y) (so the two parameters are x and y). Later it appear that the surface is instead of the form {f1(s,t),f2(s,t),f3(s,t)} with {s,t} being the parameters.

As a second remark, it seems that you have a gradient and want to recover a potential function? Existence of such a function requires that some second derivative equations be satisfied.

It could be that the problem is to:

1) First, calculate a scalar potential field,f, from the vector force field. (The statement seems to indicate there is one.)

2) Then, as a second step, find the maximum of the potential on the surface specified.

Maybe this has somehow all gotten mashed together.

A clearer statement of the objective is necessary.

If you put x3 in terms of x1 and x2 and applied the gradient you still would not have the expression first posted (for one thing, the first element of the vector x cannot be x1 after the gradient since that was the value before the gradient). Also if you knew the equation for x then why would post a questionably formulated gradient and as what it was (this is not an arcane puzzle forum)? Also Grad[ x ] evaluates without substitutions. But that doesn't matter.

When working with gradients during a calculus course it is easy to apply the Grad operator and get a nasty expressions perhaps hundreds of terms long (many are surprised by this). The answer is not to do it that way. If you are led to a wild expression in your vectors you should do as the book proposes one step at a time to maintain a solvable expressions. You will learn other methods as you advance in Mathematics but you will have to wait. Therefore technique of expressing one vector in terms of another is not "always useful".