Question

For the solid S described, do the following:

(a) Sketch the base of S in the xy-plane.
(b) Sketch a three-dimensional picture of S with the xy-plane as the floor.
(c) Compute the volume of S.

1. The base of S is the region lying above the parabola y = x 2 and below the line y = 1 over the interval 0 ≤ x ≤ 1. Cross-sections perpendicular to the x-axis are square

Answer 1

(a) and (b) see pictures attached

(c) V = 16/35

Explanation:

(a) Sketch the base of S in the xy-plane.

See picture 1 attached

(b) Sketch a three-dimensional picture of S with the xy-plane as the floor.

See picture 2 attached

(c) Compute the volume of S.

The volume is given by the triple integral

`\displaystyle\iiint_(S)zdzdydx`

The cross-sections perpendicular to the x-axis are squares so

`z=1-x^2`

The region S is given by the following inequalities

`0\leq x\leq 1\\\\x^2\leq y\leq 1\\\\0\leq z\leq 1-x^2`

Therefore

`\displaystyle\iiint_(S)zdzdydx=\displaystyle\int_(0)^(1) \displaystyle\int_(x^2)^(1) \displaystyle\int_(0)^(1-x^2) (1-x^2)dzdydx=\\\\\displaystyle\int_(0)^(1)(1-x^2)(1-x^2)(1-x^2)dx=\displaystyle\int_(0)^(1)(1-x^2)^3dx=\displaystyle(16)/(35)`

So the volume V of the solid S is

V=16/35