Let A be the region bounded by the curves y = x^2-6x + 8 and y = 0.
Find the volume obtained when A is revolved around the Y-AXIS
To find the volume obtained when the region A is revolved around the y-axis, we can use the method of cylindrical shells. Here are the steps to solve the problem:
1. Start by sketching the two curves y = x^2 - 6x + 8 and y = 0 on a coordinate plane. This will help us visualize the region A.
2. Find the intersection points of the two curves. Set the two equations equal to each other and solve for x:
x^2 - 6x + 8 = 0
Factorize or use the quadratic formula to solve for x. In this case, the equation can be factored as (x - 2)(x - 4) = 0, which gives the solutions x = 2 and x = 4.
So the region A is bounded by x = 2 and x = 4.
3. Set up the integral for the volume of the region A. We want to integrate the infinitesimally thin cylindrical shells formed by revolving the region A around the y-axis. The volume element of each shell will be given by:
dV = 2πr * h * dx
In this case, the radius r of each shell is the distance from the y-axis to the curve y = x^2 - 6x + 8, which is simply x. The height h of each shell is given by the difference in y-values between the two curves y = x^2 - 6x + 8 and y = 0. So h = (x^2 - 6x + 8) - 0 = x^2 - 6x + 8.
Therefore, the integral for the volume is:
V = ∫[2, 4] 2πx * (x^2 - 6x + 8) dx
4. Evaluate the integral. Integrate each term of the polynomial expression and evaluate it between the limits of integration 2 and 4. The integral simplifies to the following:
V = ∫[2, 4] (2πx^3 - 12πx^2 + 16πx) dx
V = [πx^4/2 - 4πx^3 + 8πx^2] evaluated from x = 2 to x = 4
V = (π * 4^4/2 - 4π * 4^3 + 8π * 4^2) - (π * 2^4/2 - 4π * 2^3 + 8π * 2^2)
V = (32π - 256π + 128π) - (2π - 32π + 32π)
V = -112π
Therefore, the volume obtained when the region A is revolved around the Y-axis is -112π cubic units.