Answer
$\dfrac{\partial w}{\partial r} = 2\pi$ and $\dfrac{\partial w}{\partial \theta} =-2\pi$
Work Step by Step
Here, we have $\dfrac{\partial w}{\partial x} = y + z$ and $\dfrac{\partial w}{\partial y} = x + z$; $\dfrac{\partial w}{\partial z} = y + x$;
Now, $\dfrac{\partial x}{\partial r} = \cos(\theta)$;
$\dfrac{\partial x}{\partial \theta} =-r \sin (\theta)$;
$\dfrac{\partial y}{\partial r} = \sin (\theta)$ ;$\dfrac{\partial y}{\partial \theta} =rcos (\theta)$
and $\dfrac{\partial z}{\partial r} = \theta$ ; $\dfrac{\partial z}{\partial \theta} = r$
Therefore:
$\dfrac{\partial w}{\partial r} =(y + z)(\cos \theta) + (x + z)(sin \space \theta) + (y + x)(\theta)$ and $\dfrac{\partial w}{\partial \theta} = (y+z)(-rsin \space \theta) + (x+z)(rcos \theta) + (y + x)(r)$
Now, find the value for each partial derivative at the given points.
Thus, we have $x =0$; $y = (2) \times \sin(\pi/2) = 2$
$z =2 \times (\pi/2) = \pi$
Hence, we get $\dfrac{\partial w}{\partial r} = \pi + \pi = 2\pi$ $\dfrac{\partial w}{\partial \theta} = -4 - 2 \times \pi +4 = -2\pi$
