Chapter 8.3, Problem 30E

a.

To determine

To find : the volume of the solid.

Answer to Problem 30E

The volume of the solid is $\frac{2\pi }{3}$ .

Explanation of Solution

Given information : The triangular region bounded by the lines $y=2x,y=0\&x=1$ about the line $x=1$ .

Calculation :

The volume of the solid:

  $\begin{array}{l}V=\pi {\displaystyle \underset{c}{\overset{d}{\int }}{\left(r\left(y\right)\right)}^{2}dy}\\ V=\pi {\displaystyle \underset{0}{\overset{2}{\int }}{\left(1-\frac{y}{2}\right)}^{2}dy}\\ V=\pi {\left[y-\frac{y^2}{2}+\frac{y^3}{12}\right]}_0^2\\ V=\pi \left[2-\frac{4}{2}+\frac{8}{12}\right]\\ \Rightarrow \frac{2\pi }{3}\end{array}$

b.

To determine

To find : the volume of the solid.

Answer to Problem 30E

The volume of the solid is $\frac{8\pi }{3}$ .

Explanation of Solution

Given information : The triangular region bounded by the lines $y=2x,y=0\&x=1$ about the line $x=2$ .

Calculation :

The volume of the solid:

  $\begin{array}{l}V=\pi {\displaystyle \underset{c}{\overset{d}{\int }}\left({\left[R\left(y\right)\right]}^2-{\left[r\left(y\right)\right]}^2\right)dy}\\ V=\pi {\displaystyle \underset{0}{\overset{2}{\int }}{\left(2-\frac{y}{2}\right)}^2-1^{2}dy}\\ V=\pi {\left[3y-y^2+\frac{y^3}{12}\right]}_0^2\\ V=\pi \left[6-4+\frac{8}{12}\right]\\ \Rightarrow \frac{8\pi }{3}\end{array}$