Find the volume of the area between xy=4, x+y=5, rotated about y=0.
In order to solve this question we will require the use of Calculus with particular reference to Washer's method.
Since we are rotating about the line y=0, the line y=0 is the same as x-axis in a two dimensional plot.
Hence, the equation we will use to find the volume between the area of the aforementioned graph is as follows:
`V =int_a^b pi*(f(x))^2 - pi*(g(x))^2 dx.`
Since we are given two equations and no limits. We need to determine the limits of these equations. These limits are found by equating the two equations.
`xy = 4=> y=4/x` (Eq.1)
`x+y=5=> y=5-x` (eq.2)
Now equate Eq.1 and Eq.2:
`4/x = 5-x`
`(4/x)*x = x(5-x)` - multiply 'x' on both sides
`4 = 5x - x^2`
`x^2 - 5x + 4 = 0` -rearrange the equation
`(x - 4)(x-1) = 0` - factorisation
`x = 4 or x=1`
Now we have our limits, b=4 and a=1. we can find the volume:
`V =int_1^4 pi*(5-x)^2 - pi*(4/x)^2 dx.`
`V =int_1^4 pi*(25 -10x + x^2) - pi*(16/x^2) dx.`
`V =int_1^4pi * (25 - 10x +x^2 - (16/x^2))dx`
`V = [pi* (25x -5x^2 +(x^3)/3 + 16/x)]` upper limit = 4, lower limit =1
Now we can substitute our limits and solve:
`V = [pi* (25(4) - 5(4)^2 + (4^3)/3 + 16/4] - [pi*(25(1) - 5(1)^2 + (1^3)/3 + 16/1)]`
`V = (136 Pi)/3 - (109pi)/3 = 9pi units^3`
Answer: The volume is cubic units or 28.27 cubic units