please solve this calc III problem?

The bounds of integration are z = 0 to z = 9 - x² - y²
where x and y are bounded inside the upper half of x² + y² = 9.

Converting to cylindrical coordinates:
z = 9 - x² - y² ==> z = 9 - r².
x² + y² = 9 ==> r = 3 (we also take θ in [0, π], since we want the upper half of the circle).

So, ∫∫∫ √(x² + y²) dV
= ∫(θ = 0 to π) ∫(r = 0 to 3) ∫(z = 0 to 9 - r²) r * (r dz dr dθ)
= π ∫(r = 0 to 3) r²z {for z = 0 to 9 - r²} dr
= π ∫(r = 0 to 3) r²(9 - r²) dr
= π ∫(r = 0 to 3) (9r² - r⁴) dr
= π(3r³ - r⁵/5) {for r = 0 to 3}
= 162π/5.

I hope this helps!

3x=4