2 Answers | Add Yours
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`
Using quadratic formula yields:
`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
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 + 2(1/x) = 7`
Multiply both sides by x to simplify.
Express the equation in quadratic form `ax^2+bx+c=0` to be able to factor.
Set each factor to zero and solve for x.
`x-2=0` and `3x-1=0`
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.
We’ve answered 315,685 questions. We can answer yours, too.Ask a question