If $\nabla u$, where u is a scalar function, =$2r^4 \vec r$, what is u and how to find it?

Question

I broke $\nabla u$ into its components (x, y and z) and evaluated each component and then integrated them, but when I add them it gives the wrong answer. I don't know much about integrating with partial derivatives.

Answer 1

Here's one way you can do it. \begin{align} \nabla u = 2r^4 \vec r &= 2r^4(x,y,z) \\ &= (2xr^4,\ 2yr^4,\ 2zr^4) \end{align} Let's do the case for the first component. We have \begin{align} u(x,y,z) = \int 2xr^4 \, dx = \int 2x(x^2+y^2+z^2)^2 \,dx \\ \end{align} When integrating with respect to one variable, we can treat all the other variables as constants ($y$ and $z$ in this case). This integral can be done using $u$-substitution: \begin{align} \textrm{Let}\ u = x^2 + y^2 + z^2 \implies du = 2x\,dx \end{align} So we get \begin{align} \int u^2 \,du = \frac{1}{3}u^3 = \frac{1}{3} (x^2+y^2+z^2)^3 + f(y,z) \end{align} where $f(y,z)$ is our arbitrary "constant" of integration, which includes possible dependence on $y$ and $z$ since they are being treated as constants.

By the symmetry of the problem, the integrals for the second and third components yield similar results: \begin{align} u(x,y,z) &= \frac{1}{3} (x^2+y^2+z^2)^3 + g(x,z) \\ u(x,y,z) &= \frac{1}{3} (x^2+y^2+z^2)^3 + h(x,y) \end{align} Hence all three of these results must be equal to each other. Canceling the common term in all of them, we see that $$ f(y,z) = g(x,z) = h(x,y) $$ The only way these functions of differing independent variables could be equal to each other for all possible choices for $x$, $y$, and $z$ is if they are all constant functions. So $$ f(y,z) = g(x,z) = h(x,y) = C $$ Giving the final result $$ u(x,y,z) = \frac{1}{3} (x^2+y^2+z^2)^3 + C $$ You can check it by taking the gradient.