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\frac zx,\frac 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))\\abla 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.