Find the volume of the solid by rotating the region bounded by the given curves about the x-axis. Use the method of cylindrical shells.
y=4x^2
2x+y=6
I drew the picture and the intersection points are at x=-1.5 and x=1. I'm not sure how to identify the "shell" with this particular graph. Help please?
Thanks.
To find the volume of the solid using the method of cylindrical shells, we need to slice the region bounded by the curves into thin cylindrical shells that are perpendicular to the x-axis. The volume of each shell is then given by the product of its height, circumference, and thickness.
In this case, the curves are y = 4x^2 and 2x + y = 6. Let's start by finding the intersection points between these curves. To do that, we can set them equal to each other:
4x^2 = 6 - 2x
Rearranging the equation, we get:
4x^2 + 2x - 6 = 0
Now we can solve this quadratic equation for x. Using the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values a = 4, b = 2, and c = -6, we get:
x = (-2 ± √(2^2 - 4(4)(-6))) / (2(4))
Simplifying further:
x = (-2 ± √(4 + 96)) / 8
x = (-2 ± √100) / 8
x = (-2 ± 10) / 8
This gives us two possible x-values:
x = (-2 + 10) / 8 = 8/8 = 1
x = (-2 - 10) / 8 = -12/8 = -3/2
So the intersection points are x = 1 and x = -3/2.
Now that we have the x-values of the intersection points, we can set up the integral to find the volume using the cylindrical shells method. The general formula for the volume of a cylindrical shell is:
V = 2πrhΔx
Where h is the height of the shell, r is the distance from the axis of rotation to the shell, Δx is the thickness of the shell, and we multiply by 2π to get the circumference.
In this case, the axis of rotation is the x-axis, so r = x.
To set up the integral, we need to express h and Δx in terms of x. Looking at the curves, the top curve is y = 4x^2 and the bottom curve is 2x + y = 6.
For a given x, the height of the shell h is given by the difference between these two curves:
h = (4x^2) - (2x + 6) = 4x^2 - 2x -6
The thickness of the shell Δx is simply a small change in x. We can express this as dx.
Now we can write the integral:
V = ∫(2πrh)dx
V = ∫(2π(x)(4x^2 - 2x -6))dx
To evaluate this integral, we need to determine the limits of integration. Since the curves intersect at x = -3/2 and x = 1, these will be the limits of integration:
V = ∫[from -3/2 to 1](2π(x)(4x^2 - 2x -6))dx
Now we can integrate the expression and evaluate the definite integral to find the volume of the solid.