Chapter 6.3, Problem 14E

To determine

Find the volume enclosed by the given curves.

Answer to Problem 14E

The volume of the solid is $\frac{32}{15}\pi$

Explanation of Solution

Given information: Given curves $y=\sqrt{x},y=0,x=1,x=-1$

Calculation:

When rotate a thin horizontal strip as shown in figure, about the $y-axis$ .

Get a washer with inner radius $y^2+1$ and outer radius is $2$ .

Note: $y=\sqrt{x}$ same as $x=y^2$ ,width of washer is $dy$ . So,

  $\begin{array}{l}dV=\pi {\left(\text{outerRadius}\right)}^2\left(\text{width}\right)-\pi {\left(\text{innerRadius}\right)}^2\left(\text{width}\right)\\ dV=\pi {\left(\text{2}\right)}^2\left(\text{dy}\right)-\pi {\left({\text{y}}^2+1\right)}^2\left(\text{dy}\right)\\ dV=4\pi \left(\text{dy}\right)-\pi \left({\text{y}}^4+2{\text{y}}^2+1\right)\left(\text{dy}\right)\\ dV=\pi \left(4-{\text{y}}^4-2{\text{y}}^2+1\right)\left(\text{dy}\right)\\ dV=\pi \left(3-{\text{y}}^4-2{\text{y}}^2\right)\left(\text{dy}\right)\end{array}$

  

To the complete volume, to integrate from $0\text{ to }1$

  $\begin{array}{l}V=\pi {\displaystyle \underset{0}{\overset{1}{\int }}\left(3-{\text{y}}^4-2{\text{y}}^2\right)}dy\\ =\pi {\left[3y-{\frac{y}{5}}^5-2{\frac{y}{3}}^3\right]}_0^1\\ =\pi \left(3-\frac{1}{5}-\frac{2}{3}\right)\\ =\pi \left(\frac{45-3-10}{15}\right)\\ =\frac{32}{15}\pi \end{array}$

Thus, the volume of the solid is $\frac{32}{15}\pi$