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....

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