Find the volume of the solid obtained by rotating the region bounded by the curves y=4x-x^2, y=8x-2x^2 about the line x=-2.
I got 256 pi/3 but pretty sure my work is wrong.
straightforward shell problem -- each shell has radius R = x + 2
and height h = 8x - 2x² - (4x - x²) = 4x - x²,
and the range is 0 ≤ x ≤ 4, so
V = ∫[a,b] 2pi Rh dx = 2pi ∫[0,4] (x + 2)(4x - x²) dx = 2pi ∫[0,4] (2x² - x^3 + 8x) dx
V = 2pi ((2/3)x^3 - (1/4)x^4 + 4x²) |[0,4] = 2pi (128/3 - 64 + 64) = 256pi/3
barring computational error.
7 months ago
To find the volume of the solid obtained by rotating the region bounded by the curves about the line x = -2, you can use the method of cylindrical shells.
First, let's graph the curves y = 4x - x^2 and y = 8x - 2x^2 to see the region of rotation:
The region of rotation lies between the two curves and is bound by the line x = -2.
To calculate the volume using cylindrical shells, we need to integrate the circumference of each cylindrical shell multiplied by its height, and sum up these volumes over the region of rotation.
Let's set up the integral:
V = ∫[a,b] 2πr(x)h(x) dx
Where r(x) is the distance from the given curve y = 4x - x^2 to the axis of rotation (x = -2), and h(x) is the height of the shell.
For each x, r(x) will be equal to the distance from the curve y = 4x - x^2 to the line x = -2 which is just 2 units.
The height of the shell, h(x), is given by the difference between the y-values of the upper and lower curves:
h(x) = (8x - 2x^2) - (4x - x^2)
h(x) = 4x - x^2
We need to find the interval of integration [a, b] where these two curves intersect.
Setting the equations equal to each other:
4x - x^2 = 8x - 2x^2
Rearranging and solving for x:
2x^2 - 4x = 0
Factor out x:
x(2x - 4) = 0
So, x = 0 or x = 2.
Therefore, the interval of integration is [0, 2].
Now, let's calculate the volume:
V = ∫[0,2] 2π(2)(4x - x^2) dx
V = 2π ∫[0,2] (8x - 2x^2) dx
V = 2π [4x^2 - (2/3)x^3] | from 0 to 2
V = 2π [(4(2)^2 - (2/3)(2)^3) - (4(0)^2 - (2/3)(0)^3)]
V = 2π [(16 - 16/3) - 0]
V = 2π [16/3]
V = (32/3)π
So, the correct volume of the solid obtained by rotating the region bounded by the curves y = 4x - x^2 and y = 8x - 2x^2 about the line x = -2 is (32/3)π.
To find the volume of the solid obtained by rotating the given region around the line x = -2, we can use the method of cylindrical shells.
First, let's find the points where the two curves intersect. Setting the two equations equal to each other, we have:
4x - x^2 = 8x - 2x^2
Simplifying, we get:
x^2 - 4x = 0
Factoring out x, we have:
x(x - 4) = 0
This gives us two possible x-values: x = 0 and x = 4. These are the x-coordinates of the points of intersection.
Next, we need to find the radius of each cylindrical shell. This is the distance between the line x = -2 and the curves y = 4x - x^2 and y = 8x - 2x^2. For this problem, we only need to consider the region between x = 0 and x = 4.
The radius (r) can be found by subtracting the x-coordinate of the point on the curve from the x-coordinate of the line x = -2:
r = (-2) - x
Next, we need to find the height (h) of each cylindrical shell. This is the difference between the y-coordinates of the two curves at a given x-value. So, we have:
h = (8x - 2x^2) - (4x - x^2)
Simplifying, we get:
h = 4x - x^2
Now, we can write the volume of each cylindrical shell as:
dV = 2πrh dx
where dx represents an infinitesimally small thickness along the x-axis. Therefore, the total volume can be expressed as:
V = ∫(2πrh) dx
To calculate this integral, we need to evaluate it from x = 0 to x = 4:
V = ∫[0 to 4](2π(4x - x^2)(-2 - x)) dx
Simplifying, we get:
V = ∫[0 to 4](2π(-8x^2 + 12x)) dx
V = -8π ∫[0 to 4](x^2 - 1.5x) dx
Evaluating the integral, we get:
V = -8π[(x^3/3) - (1.5x^2/2)] [0 to 4]
V = -8π[(64/3) - (24)]
V = -8π(64/3 - 72/3)
V = -8π(-8/3)
V = 64π/3
Therefore, the correct volume is 64π/3, not 256π/3.