Answer
$-\ln 2$
Work Step by Step
Since, $\nabla f =\dfrac{\partial f}{ \partial x} i+\dfrac{\partial f}{ \partial y}j +\dfrac{\partial f}{ \partial z} k ...(1)$
We have: $ \dfrac{\partial f}{ \partial x}= 2x \ln y -yz ; \dfrac{\partial f}{ \partial y}=\dfrac{x^2}{y} -xz \\ \dfrac{\partial f}{ \partial z}=-xy$
Now, $f=x^2\ln y -xyz +g(y,z) ...(2)$
$\implies \dfrac{\partial g(y,z)}{ \partial x}=0 $
Equation (2) becomes: $f=x^2\ln y -xyz+h(z) .....(3)$
and $h(z)=C=0$
Substitute all the above values in the given integral as follows:
$ \int_{1,2,1}^{2,1,1} (2 x \ln y ) dx +(\dfrac{x^2}{y} -xz) dy -xy dz=f(2,1,1)-f(1,2,1)=-\ln 2$
