Chapter 9, Problem 57RE

a.

To determine

To write : The volume of the solid as an improper integral.

Answer to Problem 57RE

The volume of a solid as an improper integral is $V=\pi {\displaystyle {\int }_0^{\infty }\frac{1}{{\left(y+1\right)}^2}}dy$ .

Explanation of Solution

Given information :

The infinite region in the first quadrant bounded by the coordinate axes and the curve $y=\frac{1}{x}-1$ is revolved about y-axis to generate a solid.

Calculation :

Since, the region is in first quadrant and is revolved about the y-axis to generate a solid. Therefore,

Area, $A=\pi r^2$

Also, it is rotated about the y-axis. Thus, represent the equation in terms of y.

  $\begin{array}{l}y=\frac{1}{x}-1\\ y+1=\frac{1}{x}\\ x=\frac{1}{y+1}\end{array}$

The radius will be x, so the area will be

  $A=\pi \frac{1}{{\left(y+1\right)}^2}$

The integral form 0 to infinity represents the volume. Therefore, volume

  $V=\pi {\displaystyle {\int }_0^{\infty }\frac{1}{{\left(y+1\right)}^2}}dy$

Hence,

The volume of a solid as an improper integral is $V=\pi {\displaystyle {\int }_0^{\infty }\frac{1}{{\left(y+1\right)}^2}}dy$ .

b.

To determine

To express : The integral in part (a) as a limit of a definite integral.

Answer to Problem 57RE

The improper integral as a limit of a definite integral is $V=\pi \underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{1}{{\left(y+1\right)}^2}}dy$ .

Explanation of Solution

Given information :

The infinite region in the first quadrant bounded by the coordinate axes and the curve $y=\frac{1}{x}-1$ is revolved about y-axis to generate a solid.

Calculation :

From part (a) volume of a solid as an improper integral is $V=\pi {\displaystyle {\int }_0^{\infty }\frac{1}{{\left(y+1\right)}^2}}dy$ .

Since, the above volume function is continuous is $\left[0,\infty \right)$ . Therefore, the improper integral as a limit of definite integral is

  $V=\pi \underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{1}{{\left(y+1\right)}^2}}dy$

Hence,

The improper integral as a limit of a definite integral is $V=\pi \underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{1}{{\left(y+1\right)}^2}}dy$ .

c.

To determine

To find : The volume of the solid.

Answer to Problem 57RE

The volume of the solid is $\pi$ cubic unit.

Explanation of Solution

Given information :

The infinite region in the first quadrant bounded by the coordinate axes and the curve $y=\frac{1}{x}-1$ is revolved about y-axis to generate a solid.

Calculation :

From (b) the improper integral as a limit of a definite integral is

  $V=\pi \underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{1}{{\left(y+1\right)}^2}}dy$

Solve the above equation to find the volume:

  $\begin{array}{l}V=\pi \underset{a\to \infty }{\lim }{\displaystyle {\int }_0^a\frac{1}{{\left(y+1\right)}^2}}dy\\ V=\pi \underset{a\to \infty }{\lim }{\left[-\frac{1}{\left(y+1\right)}\right]}_0^a\\ V=\pi \underset{a\to \infty }{\lim }\left[-\frac{1}{\left(a+1\right)}+\frac{1}{\left(0+1\right)}\right]\\ V=\pi \left[-0+1\right]\\ V=\pi \end{array}$

Hence,

The volume of the solid is $\pi$ cubic unit.