# Find the volume generated by revolving about the line y = 6 the area bounded by y = 1/x and 3x + 2y = 7.

lemjay | Certified Educator

To start, we need to determine the area bounded by the given equations. To do so, plot the two equations.

As shown below, the graph of y=1/x is the red curve and the graph of 3x+2y=7 is the blue line.

Hence, the bounded region is the graph at the right.

Then, solve the intersection between the blue line and the red curve using substitution method.

So, substitute y=1/x to 3x+2y=7.

`3x+2y=7`

`3x + 2(1/x) = 7`

Multiply both sides by x to simplify.

`x(3x+2/x)=7x`

`3x^2+2=7x`

Express the equation in quadratic form `ax^2+bx+c=0` to be able to factor.

`3x^2-7x+2=0`

`(x-2)(3x-1)=0`

Set each factor to zero and solve for x.

`x-2=0`                and                `3x-1=0`

`x=2`                                            `x=1/3`

Substitute values of x to y=1/x.

`x=1/3 ` ,        `y=1/(1/3) = 3`

`x=2` ,        `y=1/2`

Hence the intersection between the two equations are (1/3,3) and (2,1/2).

Next, apply the formula of disk method to determine the volume which is:

`V = pi int_a^b (r_o^2 - r_i^2) dx`

The limits of the integral a and b are the x-coordinates of the intersection of the red curve and the blue line.

So, a=1/3 and b=2.

`r_o` and `r_i` are the outer and inner radius of the disk formed when the bounded region is revolved around a certain line/axis. So radius is the distance between the axis of rotation and the curve.

To determine `r_o` and `r_i` , let's plot the bounded region and the axis of rotation which is y=6.

Base on the graph above, the outer radius is the difference between y=6 and y=1/x.

`r_o = 6 - 1/x`

Then, take the square of `r_o` .

`r_o^2= (6-1/x)^2 = 36-12/x + 1/x^2`

And the inner radius is the difference between y=6 and y=(7-3x)/2. Note that y=(7-3x)/2 is base on the equation of the blue line which is 3x+2y=7. So,

`r_i = 6 - (7-3x)/2= 12/2 - (7-3x)/2 = (3x +5)/2`

Take the square of `r_i` .

`r_i^2 = ((3x+5)/2^2 = 9/4x^2 +15/2x+25/4`

Substitute `r_o^2` and `r_i^2` to the formula of volume.

`V= pi int_(1/3)^2 [(36-12/x + 1/x^2) - (9/4x^2+15/2x+25/4)]dx`

`V= pi int_(1/3)^2 ( 119/4 -12/x+1/x^2 - 9/4x^2 - 15/2x)dx`

`V= pi (119/4x -12lnx -1/x -3/4x^3 -15/4x^2)`  `| _(1/3)^2`

`V = pi (1135/36 - 12ln6) = 31.5`

Hence, volume is 31.5 cubic units.

sciencesolve | Certified Educator

You may use the following formula to determine the volume of the solid generated revolving the region bounded by the curves `y=1/x`  and the line `3x+2y=7`  around the line y=6 such that:

`V = int_a^b pi(R^2(y) - r^2(y))dy`

You need to find the limits of integration a and b substituting `1/y`  for x in equation `3x+2y=7`  such that:

`3/y + 2y = 7 => 3 + 2y^2 - 7y = 0`

`2y^2 - 7y + 3 = 0`

`y_(1,2) = (7+-sqrt(49 - 24))/4 => y_(1,2) = (7+-sqrt25)/4`

`y_(1,2) = (7+-5)/4 => y_1 = 3 ; y_2 = 1/2`

You need to determine the radii of the washer such that:

`R(y) = 6 and r(y) = 6 - 1/` y

`V = int_(1/2)^3 pi*(36 - (6 - 1/y)^2)dy`

`V = pi int_(1/2)^3(36 - 36 + 12/y - 1/y^2)dy`

`V = pi int_(1/2)^3 12/y dy - pi int_(1/2)^3 1/y^2 dy`

`V = 12pi*ln y|_(1/2)^3+ pi/y|_(1/2)^3`

`V = 12pi(ln 3 - ln(1/2)) + pi/3 - 2pi`

`V = 12pi*ln6 - 5pi/3 => V = pi(36ln6 - 5)/3`

Hence, evaluating the volume of solid of revolution, under the given conditions, yields `V = pi(36ln6 - 5)/3` .

You may evaluate the volume such that:

Using the fundamental theorem of calculus yields:

Hence, evaluating the volume of solid generated by revolving about the line   the area bounded by   and   yields