the derivative of a 3 variable function

Question

i don't know if it's correct but as i remember, the integral of $\color{red}{df=dx + dy+dz} $ is: $\color{red}{ f=x+y+z}$ and the differential of $\color{orange}{f } $ is $\color{orange}{d(f) = {\partial f \over \partial x }dx + {\partial f \over \partial y }dy + {\partial f \over \partial z }dz }$ $\,\,$ so according to the red equation and orange equation, the integral of $d(f)$ should be $\color{green}{f={\partial f \over \partial x } + {\partial f \over \partial y } + {\partial f \over \partial z }}$. if we take $f=xyz^2$, with the orange equation df would be $df= yz^2dx + xz^2dy+2xyzdz $ and by the red equation, the f would be $\color {blue}{f=xyz^2+xyz^2 + xyz^2 }$ and it is wrong.. i'm super confused..

Answer 1

The red one is correct (except that you forgot the constant of integration), since it can be written as $$ df = d(x+y+z) , $$ which means that $f(x,y,z)=x+y+z+C$.

But the green one is wrong, since the orange one is not (in general) the same thing as $$ df = d\left( \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z} \right) . $$

Answer 2

Integration works a little differently with functions of several variables. You can only integrate with respect to one variable at a time. So, if $f(x, y, z) = x + y + z,$ and the total derivative (differential) here is just the 1-form $$df = dx + dy + dz.$$

But, recovering $f$ is a little different. We cannot just simply integrate, and integrating one at a time doesn't quite do the job, as, for example,

$$ \int dx = x + g(y, z) $$

where $g$ is a function of $y$ and $z$ only. This function serves as the "constant" of integration, as taking the derivative with respect to $x$ will make $g$ vanish.

Answer 3

The equation $$\color{orange}{ df = \frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy + \frac{\partial f}{\partial z}\,dz }\tag1$$ is true in general; the equation $$\color{red}{df = dx + dy + dz}\tag2$$ is a particular special case of Equation $(1)$ in which $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 1.$ And indeed one of the antiderivatives of Equation $(2)$ is $\color{red}{f = x + y + z}.$

It is not clear what the green equation, $\color{green}{ f \stackrel?= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z}},$ was intended to mean. What it actually does say is easy to disprove; for example, when $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 1$ the green equation says that $$ f \stackrel?= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} + \frac{\partial f}{\partial z} = 1 + 1 + 1 = 3, $$ whereas the actual solution for $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 1$ is $f = x + y + z + C,$ not the constant function $f = 3.$ What I suppose you really meant was $$ f \stackrel?= \int\frac{\partial f}{\partial x}\,dx + \int\frac{\partial f}{\partial y}\,dy + \int\frac{\partial f}{\partial z}\,dz,\tag3 $$ since that produces your blue equation, but Equation $(3)$ also is easy to disprove in the general case (as you have just done). It is easy enough to show that Equation $(3)$ is true is when $f(x,y,z) = f_1(x) + f_2(y) + f_3(z),$ but in the general case we cannot count on being able to express $f$ as a sum of three independent components in that way.

But here is how we might find $f$ in the case where $df= yz^2 \,dx + xz^2 \,dy + 2xyz \,dz$ as in your example; that is, where $\frac{\partial f}{\partial x} = yz^2,$ $\frac{\partial f}{\partial y} = xz^2,$ and $\frac{\partial f}{\partial z} = 2xyz,$ by integrating over each of the variables $x,$ $y,$ and $z.$ We do this by finding the value of $f$ at a particular point, $f(x_1,y_1,z_1).$

First, setting $g(x) = f(x,0,0),$ we find that $g'(x) = \left.\frac{\partial f}{\partial x}\right|_{y=z=0} = 0,$ and $$ g(x_1) = f(0,0,0) + \int_0^{x_1} g'(x) \,dx = f(0,0,0) + \int_0^{x_1} 0 \,dx = f(0,0,0). $$ Therefore $f(x_1,0,0) = g(x_1) = f(0,0,0).$

Next, setting $h(y) = f(x_1,y,0),$ we find that $h'(y) = \left.\frac{\partial f}{\partial y}\right|_{x=x_1,z=0} = 0,$ and $$ h(y_1) = f(x_1,0,0) + \int_0^{y_1} h'(y) \,dy = f(x_1,0,0) + \int_0^{y_1} 0 \,dx = f(x_1,0,0) = f(0,0,0). $$ Therefore $f(x_1,y_1,0) = h(y_1) = f(0,0,0).$

Finally, setting $k(z) = f(x_1,y_1,z),$ we find that $k'(z) = \left.\frac{\partial f}{\partial z}\right|_{x=x_1,y=y_1} = 2x_1y_1z,$ and \begin{align} k(z_1) &= f(x_1,y_1,0) + \int_0^{z_1} k'(z) \,dz\\ &= f(x_1,y_1,0) + \int_0^{z_1} 2x_1y_1z \,dz\\ &= f(x_1,y_1,0) + x_1y_1z_1^2 \\ &= f(0,0,0) + x_1y_1z_1^2. \end{align} Therefore $f(x_1,y_1,z_1) = k(z_1) = f(0,0,0) + x_1y_1z_1^2.$ We can treat $f(0,0,0)$ as the constant of integration, and obtain the particular antiderivative $f(x,y,z) = xyz^2$ from this result by setting $f(0,0,0) = 0.$

You could also get the same result in the same way by integrating $df$ along a path that starts at $(0,0,0),$ goes straight to $(x,0,0),$ then straight to $(x,y,0),$ and finally straight to $(x,y,z).$