Question

Let F ( x , y , z ) = ⟨ x − 1 z , y − 1 z , ln ( x y ) ⟩ . Verify that F = ∇ f , where f ( x , y , z ) = z ln ( x y ) . Evaluate ∫ C F ⋅ d r , where r ( t ) = ⟨ e t , e 2 t , t 2 ⟩ for 1 ≤ t ≤ 3 . Evaluate ∫ C F ⋅ d r for any path C from P = ( 1 2 , 4 , 2 ) to Q = ( 2 , 2 , 3 ) contained in the region x > 0 , y > 0 . In part (c), why is it necessary to specify that the path lies in the region where x and y are positive?

Answer 1

I suppose you mean

$\vec F(x,y,z)=\left\langle\dfrac zx,\dfrac zy,\ln(xy)\right\rangle$

Showing that $f(x,y,z)=z\ln(xy)$ has gradient equal to $\vec F$ is trivial.

By virtue of the existence of $f$, the gradient theorem applies here, so

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_{(12,4,2)}^{(2,2,3)}\nabla f(x,y,z)\cdot\mathrm d\vec r=f(2,2,3)-f(12,4,2)=3\ln 4-2\ln48=\boxed{-\ln36}$

The requirement that $x>0$ and $y>0$ is a consequence of the domain of the logarithm function. Both have to be positive in order for $\ln(xy)$ to exist.