Bowl of water

This was a bit messy in terms of calculations. Here is my shot at it:

Density is:

$\rho = 100000z$

Converting to spherical coordinates gives:

$x = r\sin{\theta}\cos{\phi}$ $y = r\sin{\theta}\sin{\phi}$ $z = r\cos{\theta}$

The volume element in spherical coordinates is:

$dV = r^2 \sin{\theta} \ dr \ d\theta \ d\phi$

For the bowl the mass in kilograms is:

$dM = \rho \ dV$ $\implies M_B = 10^5 \int_{0.05}^{0.05+\frac{2.5}{1000}} \int_{\pi/2}^{\pi} \int_{0}^{2 \pi} r^3 \sin{\theta}\cos{\theta} \ d\phi \ d\theta \ dr$ $M_B = \frac{34481 \pi}{1024000}$

The mass of water in kilograms is:

$M_W = 1000\left(\frac{2}{3} \pi \left(\frac{5}{100}\right)^3\right)=\frac{250 \pi}{3000}$

The answer in kilograms is the sum of $M_B$ and $M_W$ and that is converted to grams. The required answer is $\boxed{368}$ .

The given density function is strange. If one were to treat a hollow sphere of same inner and outer radius and attempt to calculate its mass, mathematically, it would evaluate to zero.