Question

Determine whether or not the vector field is conservative. If it is, find a function f such that F = ∇f. If the vector field is not conservative, enter NONE. F(x, y, z) = 5y2z3 i + 10xyz3 j + 15xy2z2 k f(x, y, z) = + K

Answer 1

$\vec F$ is conservative if we can find a scalar function $f$ such that $\nabla f=\vec F$. This would require

$\dfrac{\partial f}{\partial x}=5y^2z^3$

$\dfrac{\partial f}{\partial y}=10xyz^3$

$\dfrac{\partial f}{\partial z}=15xy^2z^2$

Integrate both sides of the first PDE with respect to $x$:

$f(x,y,z)=5xy^2z^3+g(y,z)$ (*)

Differentiate both sides of (*) with respect to $y$:

$\dfrac{\partial f}{\partial y}=10xyz^3=10xyz^3+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

Differentiate both sides of (*) with respect to $z$:

$\dfrac{\partial f}{\partial z}=15xy^2z^2=15xy^2z^2+\dfrac{\mathrm dh}{\mathrm dz}\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

So we have

$f(x,y,z)=5xy^2z^3+C$

and so $\vec F$ is indeed conservative.