Question

Let,Ę. U = 2r4r, Find U​

Answer 1

Answer:

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Step-by-step explanation:

Here's one way you can do it.

∇u=2r4r⃗ =2r4(x,y,z)=(2xr4, 2yr4, 2zr4)

Let's do the case for the first component. We have

u(x,y,z)=∫2xr4dx=∫2x(x2+y2+z2)2dx

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:

Let u=x2+y2+z2⟹du=2xdx

So we get

∫u2du=13u3=13(x2+y2+z2)3+f(y,z)

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:

u(x,y,z)u(x,y,z)=13(x2+y2+z2)3+g(x,z)=13(x2+y2+z2)3+h(x,y)

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)=13(x2+y2+z2)3+C

You can check it by taking the gradient.