Please write out all steps so I can learn Evaluate the integ

Please write out all steps so I can learn! Evaluate the integral of (x^2 + y^2)z^2 over part of the cylinder x^2 + y^2 <= 1 inside the sphere x^2 + y^2 + z^2 = 4

Solution

x²+y²+z²=4,

z=-[4 -(x²+y²)],z=[4 -(x²+y²)]

in cylindrical coordinates

x=rcos y=rsin

x²+y²=r²

z=-[4 -(r²)] , z=[4 -(r²)]

x²+y²<=1

r²<=1

0<=r<=1

0<=<=2,0<=r<=1 ,-[4 -(r²)]<=z<=[4 -(r²)]

dv =r dz dr d

   (x²+y²)z² dv

= _{[0 to 2]  [0 to 1] [-[4 -(r²)] to [4 -(r²)]]}(r²)z² r dz dr d

= _{[0 to 2]  [0 to 1] [-[4 -(r²)] to [4 -(r²)]]} (1/3)r³z³ dr d

= _{[0 to 2]  [0 to 1]} (1/3)r³[([4 -r²])³ - (-[4 -r²])³] dr d

= _{[0 to 2]  [0 to 1]} (2/3)r³([4 -r²])³ dr d

= _{[0 to 2]  [0 to 1]} (1/3)r²([4 -r²])³ 2r dr d

let 4 -r²=u=>r²=4-u =>2r dr =-du, r =0 =>u =4 , r =1=>u =3

= _{[0 to 2]  [4 to 3]} (1/3)(4-u)(u)³ (-du) d

= _{[0 to 2]  [3 to 4]} (1/3)(4-u)u^3/2du d

= _{[0 to 2]  [3 to 4]} (1/3)(4u^3/2-u^5/2)du d

= _{[0 to 2]  [3 to 4]} (1/3)(4(2/5)u^5/2-(2/7)u^7/2) d

= _{[0 to 2]  [3 to 4]} (2/3)((4/5)u^5/2-(1/7)u^7/2) d

= _{[0 to 2]}  (2/3)((4/5)4^5/2-(1/7)4^7/2) -(2/3)((4/5)3^5/2-(1/7)3^7/2) d

= [(2/3)((4/5)4^5/2-(1/7)4^7/2) -(2/3)((4/5)3^5/2-(1/7)3^7/2) ](2-0)

= [((4/5)4^5/2-(1/7)4^7/2) -((4/5)3^5/2-(1/7)3^7/2) ](4/3)

=6.385