Chapter 6.3, Problem 34E

Let T be the triangular region with vertices (0, 0), (1, 0), and (1, 2), and let V be the volume of the solid generated when T is rotated about the line x = a, where a > 1. Express a in terms of V.

To determine

To express: The limit (a) in terms of volume of solid (V) using the method of cylindrical shell.

Answer to Problem 34E

The expression of the limit (a) in terms of V is $\underset{\_}{0.159V+0.6667}$.

Explanation of Solution

Given:

The co-ordinates of the triangular region $\left(x,y\right)$ are $\left(0,0\right),\left(1,0\right)$, and $\left(1,2\right)$.

The equations are $y=2x$, $x=a;$ where $a>1$.

Calculation:

Plot a graph using the co-ordinates of the triangular region $\left(x,y\right)$ are $\left(0,0\right),\left(1,0\right),$ and $\left(1,2\right)$:

Draw the shell as shown in Figure 1.

Calculate the volume using the method of cylindrical shell:

$V={\displaystyle {\int }_a^{b}2\pi x\left[f\left(x\right)\right]dx}$ (1)

Substitute 0 for a, 1 for b, $\left(a-x\right)$ for x, and 2x for $\left[f\left(x\right)\right]$ in Equation (1).

$\begin{array}{c}V={\displaystyle {\int }_0^{1}2\pi \left(a-x\right)\left(2x\right)dx}\\ =4\pi {\displaystyle {\int }_0^1\left(ax-x^2\right)dx}\end{array}$ (2)

Integrate Equation (2).

$\begin{array}{c}V=4\pi {\left[a\left(\frac{x^{1+1}}{1+1}\right)-\left(\frac{x^{2+1}}{2+1}\right)\right]}_0^1\\ =4\pi {\left[\frac{a}{2}{x}^2-\frac{1}{3}{x}^3\right]}_0^1\\ =4\pi \left[\left(\frac{a\times 1^2}{2}-\frac{1^3}{3}\right)-0\right]\end{array}$

$\begin{array}{c}=4\pi \left(0.5a-\frac{1}{3}\right)\\ V=6.283a-4.18879\end{array}$ (3)

Rearrange Equation (3).

$\begin{array}{c}6.283a=V+4.18879\\ a=\frac{V+4.18879}{6.283}\\ =0.159V+0.6667\end{array}$

Hence, the expression of the limit (a) in terms of V is $\underset{\_}{0.159V+0.6667}$.