Question

Use the disk method or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about each given line. y=6/x^2 ,y=0,x=1,x=3 .
1) Find the y-axis
2) Find the line y=6

Answer 1

Answer:

1) V = 12 π  ㏑ 3

2) $\mathbf{V = \dfrac{328 \pi}{9}}$

Step-by-step explanation:

Given that:

the graphs of the equations about each given line is:

$y = \dfrac{6}{x^2}, y =0 , x=1 , x=3$

Using Shell method to determine the required volume,

where;

shell radius = x;   &

height of the shell = $\dfrac{6}{x^2}$

∴

Volume V = $\int ^b_{x-1} \ 2 \pi ( x) ( \dfrac{6}{x^2}) \ dx$

$V = \int ^3_{x-1} \ 2 \pi ( x) ( \dfrac{6}{x^2}) \ dx$

$V = 12 \pi \int ^3_{x-1} \dfrac{1}{x} \ dx$

$V = 12 \pi ( In \ x ) ^3_{x-1}$

V = 12 π ( ㏑ 3 - ㏑ 1)

V = 12 π ( ㏑ 3 - 0)

V = 12 π  ㏑ 3

2) Find the line y=6

Using the disk method here;

where,

Inner radius $r(x) = 6 - \dfrac{6}{x^2}$

outer radius R(x) = 6

Thus, the volume of the solid is as follows:

$V = \int ^3_{x-1} \begin {bmatrix} \pi (6)^2 - \pi ( 6 - \dfrac{6}{x^2})^2 \end {bmatrix} \ dx$

$V = \pi (6)^2 \int ^3_{x-1} \begin {bmatrix} 1 - \pi ( 1 - \dfrac{1}{x^2})^2 \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} 1 - ( 1 + \dfrac{1}{x^4}- \dfrac{2}{x^2}) \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} - \dfrac{1}{x^4}+ \dfrac{2}{x^2} \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} {-x^{-4}}+ 2x^{-2} \end {bmatrix} \ dx$

Recall that:

$\int x^n dx = \dfrac{x^n +1}{n+1}$

Then:

$V = 36 \pi ( -\dfrac{x^{-3}}{-3}+ \dfrac{2x^{-1}}{-1})^3_{x-1}$

$V = 36 \pi ( \dfrac{1}{3x^3}- \dfrac{2}{x})^3_{x-1}$

$V = 36 \pi \begin {bmatrix} ( \dfrac{1}{3(3)^3}- \dfrac{2}{3}) - ( \dfrac{1}{3(1)^3}- \dfrac{2}{1}) \end {bmatrix}$

$V = 36 \pi (\dfrac{82}{81})$

$\mathbf{V = \dfrac{328 \pi}{9}}$

The graph of equation for 1 and 2 is also attached in the file below.