Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y = x2, y = 1; about y = 6
Hint: use the disk method
Find the intersection of x² and y=6, which is x=±sqrt(6).
You may want to integrate with x=-sqrt(6) to +sqrt(6)
πf(x)^2 dx [equiv. to πr²]
where
f(x)=x²-1 [i.e. from y=1 to y=x^2]
To find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis, we can use the method of cylindrical shells.
First, let's draw a rough sketch of the region bounded by the curves y = x^2 and y = 1:
|
|
| * y = 1
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| *
| * * *
| * * * * *
--------------
| | |
| | |
| | |
| | |
| --|----|---- x-axis
From the sketch, we see that the region is bounded by two curves: y = x^2 and y = 1. To determine the limits of integration for x, we need to find the x-values where these curves intersect.
Setting the two equations equal, we have:
x^2 = 1
Taking the square root of both sides, we get:
x = ±1
So, the region is bounded by x = -1 and x = 1.
Now, let's consider a vertical slice of the region. This vertical slice has a height of (y - 6) and a width of dx.
To find the volume of each cylindrical shell, we need to find the circumference and height of each shell. The circumference, which represents the length around each shell, is given by 2πy. The height, which represents the thickness of each shell, is given by y - 6.
Since we integrated with respect to x, we need to express y in terms of x. Rearranging y = x^2, we find that x = √y.
Using the formula for the volume of a cylinder shell, the volume of each shell is given by:
dV = 2πy * (y - 6) * dx
Now, we can set up the integral to calculate the total volume of the solid:
V = ∫[from x=-1 to x=1] 2πy * (y - 6) dx
Substituting y = x^2 and dx = dy/2√y, the integral becomes:
V = ∫[from y=0 to y=1] 2πx^2 * (x^2 - 6) (dy/2√y)
V = π∫[from y=0 to y=1] x^2 * (x^2 - 6) / √y dy
Simplifying the expression inside the integral, we get:
V = π∫[from y=0 to y=1] (x^4 - 6x^2) / √y dy
Now, we can integrate with respect to y:
V = π[ (1/5)x^5 - 2x^3 ] [from y=0 to y=1]
Substituting x = √y, the limits of integration become:
V = π[ (1/5)(√y)^5 - 2(√y)^3 ] [from y=0 to y=1]
V = π[ (1/5)y^5/2 - 2y^(3/2) ] [from y=0 to y=1]
V = π[ (1/5) - 2(1) ] = π[ (1/5) - 2 ]
Therefore, the volume of the solid obtained by rotating the region bounded by y = x^2, y = 1, and y = 6 about y = 6 is π(1/5 - 2) or -9π/5.