Use the Shell Method to compute the volume V of the solid obtained by rotating the region enclosed by the graphs of the functions
y = x^2, y = 8 − x^2,and x = 1/2
about the y-axis.
Here is how I set up the integral:
2 pi integral sign [-2,2] (1/2 - x)[(8-x^2)- (x^2)]dx
Since x=1/2 is greater than a (a= -2), I subtracted x from 1/2. Is this correct?
Also, since the graph of y= 8- x^2 is above the graph of y=x^2, I subtracted x^2 from 8-x^2. Is this correct?
I got (64 pi)/3, but the answer is wrong.
Why do you integrate over [-2,2]?
The curves intersect at (-2,4) and (2,4) but what does the line x = 1/2 have to do with it? There's a region left of the line, and a region right of the line. Since we are rotating about the y-axis, I think we want the region from x=1/2 to x=2, otherwise the axis of rotation is within the solid being generated.
So, with that in mind, if what I have said is correct, then
v = ∫[1/2,2] 2πrh dx
where r = x and h=(8-x^2)-x^2
v = 2π∫[1/2,2] x(8-2x^2) dx = 225π/16
Now, the way you set up your integral makes me think you are rotating around the line x = 1/2, and not the y-axis. If so, you probably need to integrate over [1/2,2], not [-2,2] and use the factor (x-1/2) not (1/2-x).
The setup of the integral is correct, but there are a couple of errors in the expression.
First, you mentioned that x = 1/2 is greater than a = -2, so you subtracted x from 1/2. However, in the shell method, you need to use the distance from the axis of rotation, which is the y-axis in this case. Since you are rotating about the y-axis, the distance is actually x.
So instead of (1/2 - x), you should have (x - 0).
Next, when calculating the height of the shell, you correctly mentioned that the graph of y = 8 - x^2 is above the graph of y = x^2. However, when subtracting the functions to calculate the height, you should consider the function that is further from the axis of rotation minus the function that is closer.
In this case, the graph of y = 8 - x^2 is further from the y-axis than the graph of y = x^2. Therefore, you should subtract x^2 from (8 - x^2), not the other way around.
With these corrections, the integral should be:
V = 2π ∫[a,b] (x - 0) [(8 - x^2) - x^2] dx
Now, let's calculate it step-by-step:
V = 2π ∫[-2, 2] (x - 0) (8 - 2x^2) dx
V = 2π ∫[-2, 2] (8x - 2x^3) dx
To integrate this, you can use the power rule:
V = 2π [4x^2 - (1/2)x^4] |[-2, 2]
V = 2π [(4(2)^2 - (1/2)(2)^4) - (4(-2)^2 - (1/2)(-2)^4)]
V = 2π [(16 - 8) - (16 - 8)]
V = 2π [8 - 8]
V = 0
Therefore, the volume of the solid obtained by rotating the region enclosed by the given functions about the y-axis is 0.
To use the shell method to compute the volume of the solid obtained by rotating the region about the y-axis, you need to integrate the circumference of the cylindrical shells multiplied by their height.
First, let's find the limits of integration. The region is enclosed by the graphs of the functions y = x^2, y = 8 - x^2, and x = 1/2. To determine the limits of integration, we need to find the x-values where these curves intersect.
Setting y = x^2 and y = 8 - x^2 equal to each other, we get:
x^2 = 8 - x^2
2x^2 = 8
x^2 = 4
x = ±2
However, we are only interested in the region to the right of the y-axis, so the limits of integration for x are from 0 to 2.
Now let's set up the integral using the shell method:
The radius of each shell is the distance from the y-axis to an x-value, which is x.
The height of each shell is the difference between the two functions at that x-value, which is (8 - x^2) - x^2 = 8 - 2x^2.
The volume can be calculated using the formula:
V = 2π ∫ [a, b] x (8 - 2x^2) dx
So, substituting the limits of integration, the integral becomes:
V = 2π ∫ [0, 2] x (8 - 2x^2) dx
Now, let's evaluate the integral:
V = 2π ∫ [0, 2] (8x - 2x^3) dx
Integrating, we get:
V = 2π [4x^2 - (1/2)x^4] | [0, 2]
Plugging in the limits of integration, we have:
V = 2π [(4(2^2) - (1/2)(2^4)) - (4(0^2) - (1/2)(0^4))]
Simplifying:
V = 2π [(16 - 8) - (0 - 0)]
V = 2π [8]
V = 16π
So, the correct volume is 16π, which is not equal to (64π)/3.
Therefore, there may have been an error in your calculations or in the setup of the integral. Please double-check the steps and calculations to find the mistake.