Answer
$\dfrac{\partial w}{\partial r} = 2\pi$ and $\dfrac{\partial w}{\partial \theta} =-2\pi$
Work Step by Step
Here, we have $\dfrac{\partial w}{\partial x} = y + z\\ \dfrac{\partial w}{\partial y} = x + z\\\dfrac{\partial w}{\partial z} = y + x$;
Now, $\dfrac{\partial x}{\partial r} = \cos(\theta)\\\dfrac{\partial x}{\partial \theta} =-r \times \sin (\theta)\\ \dfrac{\partial y}{\partial r} = \sin (\theta)\\ \dfrac{\partial y}{\partial \theta} =r \times cos (\theta)$
$\dfrac{\partial z}{\partial r} = \theta$ ; $\dfrac{\partial z}{\partial \theta} = r$
Therefore:
$\dfrac{\partial w}{\partial r} =(y + z)(\cos \theta) + (x + z)(sin \space \theta) + (y + x)(\theta);\\\dfrac{\partial w}{\partial \theta} = (y+z) \times (-rsin \times \space \theta) + (x+z) \times (rcos \theta) + (y + x)(r)$
Now, find the value for each partial derivative at the given points.
Thus, we have $x =0 \\ y = (2)[ \sin(\pi/2) ]= 2;z =2 \times (\pi/2) = \pi$
Hence, we get $\dfrac{\partial w}{\partial r} = \pi + \pi = 2\pi$ and $\dfrac{\partial w}{\partial \theta} = -2\pi$
