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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should evaluate the volume of solid of revolution using the formula:

`V = int_a^b A(x)dx`

A(x) represents the area of cross sectional area

`A(x) = pi(R^2 - r^2)`

You need to evaluate the volume of solid of revolution, bounded by the curves `y = 1/x`  and `3x + 2y = 7 =>`  `y = (7-3x)/2` , about the line y=6.

Notice that the inner radius is `r = 6 - (7-3x)/2`  and the outer radius is `R = 6 - 1/x` , hence, evaluating the area, yields:

`A(x) = pi((6 - 1/x)^2 - (5 + 3x)^2/4)`

You need to find the limits of integration setting up the equation such that:

`1/x = (7-3x)/2 => 7x - 3x^2 - 2 = 0`

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

Using quadratic formula yields:

`x_(1,2) = (7 +- sqrt(49 - 24))/6 => x_(1,2) = (7 +- sqrt25)/6`

`x_(1,2) = (7 +- 5)/6 => x_1 = 2 ; x_2 = 1/3`

You may evaluate the volume such that:

`V = pi*int_(1/3)^2 (6 - 1/x)^2 dx - pi/4*int_(1/3)^2(5 + 3x)^2 dx`

Expanding the squares yields:

`V = pi int_(1/3)^2 (36 - 12/x + 1/x^2)dx - (pi/4)int_(1/3)^2(25 + 30x + 9x^2)dx`

`V = pi(36x - 12ln x - 1/x)|_(1/3)^2 - (pi/4)(25x + 15x^2 + 3x^3)|_(1/3)^2 `

`V = pi(72 - 12ln 2 - 1/2 - 12 + 12ln(1/3) + 3) - (pi/4)(50 + 60 + 24 - 25/3 - 5/3 - 1/9)`

`V = pi(63 - 1/2 - 12 ln 6)- 1115pi/9`

`V = 125*pi/2 - 12ln6*pi - 1115pi/9`

Hence, evaluating the volume of the solid of revolution, under the given conditions, yields `V = 125*pi/2 - 12ln6*pi - 1115pi/9.`

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