Chapter 7.3, Problem 55E

(a)

To determine

The volume of the solid by the Washer method.

Answer to Problem 55E

The volume of the solid is $\frac{11\pi }{48}$

Explanation of Solution

Given:

The region is bounded by the curve $y=\frac{1}{\sqrt{x}}$ , on the left by the line $x=\frac{1}{4}$ and the line $y=1$ .

Calculation:

The volume can be find out as:

  $\begin{array}{l}V={\displaystyle \underset{1}{\overset{2}{\int }}\pi {\left(\frac{1}{y^2}\right)}^2-{\left(\frac{1}{4}\right)}^{2}dy}\\ V=\pi {\displaystyle \underset{1}{\overset{2}{\int }}{\left(\frac{1}{y^2}\right)}^2-\left(\frac{1}{16}\right)}dy\\ V=\pi {\left(-\frac{1}{3y^2}-\frac{1}{16y}\right)}_1^2\\ V=\frac{11\pi }{48}\end{array}$

Conclusion:

The volume of the solid is $\frac{11\pi }{48}$

(b)

To determine

The volume of the solid by the Cylindrical shell method.

Answer to Problem 55E

The volume of the solid is $\frac{11\pi }{48}$

Explanation of Solution

Given:

The region is bounded by the curve $y=\frac{1}{\sqrt{x}}$ , on the left by the line $x=\frac{1}{4}$ and the line $y=1$ .

Calculation:

The volume can be find out as:

  $\begin{array}{l}V={\displaystyle \underset{1}{\overset{2}{\int }}2\pi xf(x)dx}\\ V=2\pi {\displaystyle \underset{\frac{1}{4}}{\overset{1}{\int }}x(\frac{1}{\sqrt{x}}-1})dx\\ V=2\pi {\displaystyle \underset{\frac{1}{4}}{\overset{1}{\int }}(\sqrt{x}-x)dx}\\ V=2\pi {\left[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{1}{2}{x}^2\right]}_{\frac{1}{4}}^1\\ V=\frac{11\pi }{48}\end{array}$

Conclusion:

The volume of the solid is $\frac{11\pi }{48}$