Question

To find the volume of the solid with the given description.

Short answer

The volume of the solid with the given description is \(\frac{8}{{15}}\).

Step-by-step solution

Given data

It is given that the base of the solid \(S\) is the region enclosed by the parabola \(y = 1 - {x^2}\) and the \(x\)-axis.

Concept used of Volume

Volumeis a scalar quantity expressed by the amount of three-dimensional space enclosed by a closed surface.

Solve to find the volume

The cross-sections that are perpendicular to the \(x\)-axis are isosceles triangles such that the height is equal to the base.

That is, base = height \( = 1 - {x^2}\).

As the isosceles triangle lies on the \(x\)-axis, the values of \(x\) will vary from \( - 1\) to 1 .

That is, on \(x\)-axis the height is 0 .

\(\begin{aligned}{}1 - {x^2} = 0\\{x^2} = 1\\x = \pm 1\end{aligned}\)

Obtain the area of the cross section is

\(\begin{aligned}{}A(x) = \frac{1}{2} \cdot \left( {1 - {x^2}} \right) \cdot \left( {1 - {x^2}} \right)\\ = \frac{1}{2}\left( {1 - 2{x^2} + {x^4}} \right)\end{aligned}\)

Then the volume is computed as

\(\begin{aligned}{}V = \int_{ - 1}^1 {\frac{1}{2}} \left( {1 - 2{x^2} + {x^4}} \right)dx\\ = \frac{1}{2}\left( {x - \frac{{2{x^3}}}{3} + \frac{{{x^5}}}{5}} \right)_{ - 1}^1\\ = \frac{1}{2}\left( {(1) - \frac{{2{{(1)}^3}}}{3} + \frac{{{{(1)}^5}}}{5} - ( - 1) + \frac{{2{{( - 1)}^3}}}{3} - \frac{{{{( - 1)}^5}}}{5}} \right)\\ = \frac{1}{2}\left( {1 - \frac{2}{3} + \frac{1}{5} + 1 - \frac{2}{3} + \frac{1}{5}} \right)\end{aligned}\)

On further simplifications,

\(\begin{aligned}{}V = 1 - \frac{2}{3} + \frac{1}{5}\\ = \frac{{15 - 10 + 3}}{{15}}\\ = \frac{8}{{15}}\end{aligned}\)

Therefore, the volume of the solid with the given description is \(\frac{8}{{15}}\).