# Find the volume of the area between xy=4, x+y=5, rotated about y=0.

Get ad-free questions with an eNotes 48-hour free trial.

Try It Free No Thanks
uwais17 | Certified Educator

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