Chapter 8.3, Problem 3E

To determine

To find : the volume of the solid.

Answer to Problem 3E

The volume of the solid is $16{\text{ units}}^3$ .

Explanation of Solution

Given information : The solid lies between planes perpendicular to the x -axis at $x=0\&x=4$ . The cross sections perpendicular to the axis on the interval $0\le x\le 4$ are squares whose diagonals run from $y=-\sqrt{x}\text{to }y=\sqrt{x}$ .

Calculation :

The diagonals run from $y=-\sqrt{x}\text{to }y=\sqrt{x}$ . So, the length of one diagonal is $\sqrt{x}-\left(-\sqrt{x}\right)=2\sqrt{x}$ .

If the length of a side of a square with diagonal $2\sqrt{x}$ is $s$ then the area of one square at $x$ is $s^2$ .

  $\begin{array}{l}{s}^2+s^2={\left(2\sqrt{x}\right)}^2\\ 2s^2=4x\\ s^2=2x\\ A=s^2=2x\end{array}$

Integrate from $0\text{ to }4$ to find the volume of the solid.

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{4}{\int }}2xdx}\\ V={\left[x^2\right]}_0^4\\ V=16-0\\ V=16{\text{ units}}^3\end{array}$