Question

Find the volume enclosed by the cylinder `y^2=x` and `y=x^2` Cut off by the planes z = 0, x+y+z=2.

Answer 1

The solid is bounded by the parabolas `y^2=x`, `y=x^2` in the x y plane.

In x-y-z plane x+y+z =2 is top base.
The volume between this curves is given by ,
V = ∫∫𝒛 𝒅𝒙 𝒅𝒚= ∫∫(𝟐−𝒙−𝒚)𝒅𝒙𝒅𝒚
From the diagram we can conclude that the intersection point of both Parabolas are (0,0),(1,1).

`therefore "V"=int_0^1int_(x^2)^sqrtx(2-x-y)dxdy`

`=int_0^1(2y-xy-y^2/2)_(x^2)^sqrtxdx`

`=int_0^1[(2sqrtx-xsqrtx-x/2)-(2x^2-x^3-x^4/2)]dx`

`=[(4x^(3/2))/3-(2x^(5/2))/5-x^2/4-(2x^3)/3+x^4/4+x^5/10]_0^1`

`therefore "V"=11/30`