Calculus 3 Triple Integrals

Question

Find ∫∫∫Ey dV, where E is the solid bounded by the parabolic cylinder z=x^2 and the planes y=0 and z=13−2y

Answer 1

The projection of the solid on the xy-plane is the parabola y = 13/2 −x²/2

Then the given integral is ∫^{x=13^0.5}_x=0 ∫^{y=13/2-x^2/2}_y=0 ∫^z=13-2y_z=0 y dzdydx=

∫^{x=13^0.5}_x=0 ∫^{y=13/2-x^2/2}_y=0 [yz]|₀^z=13-2ydydx =

∫^{x=13^0.5}_x=0 ∫^{y=13/2-x^2/2}_y=0[13y-2y²] dydx =

∫^{x=13^0.5}_x=0 [13y²/2− 2y³/3]| ^{y=13/2-x^2/2}_y=0dx

∫^{x=13^0.5}_x=0[ 13/2( 13/2−x²/2)²−(2/3) ( 13/2−x²/2)³]dx=

41743√13/420