integral problem

Looking at the limits of integration, as well as the integrand, I think changing to spherical coordinates would make sense. In spherical coordinates, 4−x2−y2−z2=4−ρ2\displaystyle \sqrt{4- x^2- y^2- z^2}= \sqrt{4- \rho^2} and the differential of volume is ρ2sin(θ)dρdθdϕ\displaystyle \rho^2 sin(\theta)d\rho d\theta d\phi. rho\displaystyle rho goes from 0 to 2, θ\displaystyle \theta goes from 0 to π\displaystyle \pi, and ϕ\displaystyle \phi goes from 0 to π/2\displaystyle \pi/2. The integral is ∫ϕ=0π2∫θ=0π∫ρ=024−ρ2ρ2dρdθdϕ\displaystyle \int_{\phi= 0}^{\frac{\pi}{2}}\int_{\theta= 0}^\pi \int_{\rho=0}^2 \sqrt{4- \rho^2} \rho^2 d\rho d\theta d\phi.
 
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