Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis.
1)y = 8 − x^2, y = x^2; about x = 2
v=?
Tried it a bunch, can't seem to get it right.
so, what did you do? Post your attempt.
See the other examples I did below.
To find the volume generated by rotating the region bounded by the curves y = 8 − x^2 and y = x^2 about the line x = 2, we can use the method of cylindrical shells.
The volume element of a cylindrical shell is given by the formula:
dV = 2πrh * dr,
where h is the height of the shell, r is the distance from the axis of rotation (in this case, x = 2), and dr is the thickness of the shell.
To set up the integral, we need to express r and h in terms of x.
First, let's graph the region bounded by the curves.
From y = 8 − x^2 and y = x^2, we find their intersection points. Setting the equations equal to each other, we have:
8 − x^2 = x^2
8 = 2x^2
4 = x^2
x = ±2
Therefore, the curves intersect at x = -2 and x = 2.
Now, let's calculate the height of the shell, h. The upper curve (8 − x^2) is further from the axis of rotation, so the height of the shell will be given by:
h = (8 − x^2) - (x^2) = 8 - 2x^2
Next, we need to express r in terms of x. Since the axis of rotation is x = 2, r will be given by:
r = x - 2
Finally, we need to determine the limits of integration for x. The intersection points x = -2 and x = 2 define the range of x values for the region of interest. Therefore, our limits of integration will be -2 to 2.
Now, we can set up the integral to find the volume:
V = ∫[from -2 to 2] (2πrh) dr
Substituting the expressions for h and r, our integral becomes:
V = ∫[from -2 to 2] (2π(8 - 2x^2)(x-2)) dx
Now, you can evaluate this integral to find the volume V by solving it using any suitable integration technique.
I hope this explanation helps you understand how to set up the integral using the method of cylindrical shells.