Answer
$V = 2u^{2}$
Work Step by Step
1. Graph the parabola $y = 1 - x^2$ . A cross section that represents the side of a square should be shown.
2. Let x be the side of the square.
3. Solve for x:
$y = 1 - x^2$
$x^2 = 1 - y$
$x = -\sqrt 1 - \sqrt y$
$x = \sqrt 1-\sqrt y$
4. Since a measurement is always positive, ignore the negative x-value. Only one of the values of x was considered, so it’s only half of the parabola. Therefore, double the considered x value to get a value for the entire parabola:
$x = 2(\sqrt 1-\sqrt y)$
5. Area of a square is found through the equation:
$A = x^{2}$
6. Substitute $x = 2(\sqrt (1 - y)$ into the equation:
$ A = (2(\sqrt(1-y))^2$
7. Solve for A:
$A = -4y + 4$
8. Integrate the computed area of the square to find its volume:
$V = \int(4 - 4y)dy |^{1} _{0}$
9. Solve for V:
$ = 2$
10. Thus, the volume of the described solid S is 2 cubic units.
$V = 2u^{2}$
