Answer
$\ln (\dfrac{\pi}{2})$
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}= 2 \cos y ; \dfrac{\partial f}{ \partial y}=\dfrac{1}{y}-2 x \sin y \\ \dfrac{\partial f}{ \partial z}=\dfrac{1}{z}$
Now, $f=2x \cos y +g(y,z) ...(2)$
$\implies \dfrac{\partial g(y,z)}{ \partial x}=\dfrac{1}{y}$
and $g(y,z)=\ln y + h(z)$
Equation (2) becomes: $f=2x \cos y +\ln y+h(z) .....(3)$
and $h(z)=\ln z+C$
Substitute all the above values in the given integral as follows:
$ \int_{0,2,1}^{1,0.5 \pi,2} 2 \cos y dx+(\dfrac{1}{y}-2 x \sin t ) dy +\dfrac{dz}{z}=\ln (\pi/2)+\ln 2 -\ln 2=\ln (\dfrac{\pi}{2})$
