Use the disk method to find the volume of the solid generated by revolving about the y-axis the region bounded by the curves y=8−x^2 and the curve y=x^2.
Since the y-axis is also the axis of symmetry for the curves, we can just take one side and revolve it. This will be easiest using shells of thickness dx:
v = ∫[0,2] 2πrh dx
where r=x and h=(8-x^2)-x^2=8-2x^2
v = ∫[0,2] 2πx(8-2x^2) dx = 16π
To find the volume of the solid generated by revolving the region bounded by the curves y = 8 - x^2 and y = x^2 about the y-axis, we can use the disk method.
Step 1: Identify the bounds for the y-values.
We need to find the points of intersection between the two curves. Setting the equations equal to each other, we have:
8 - x^2 = x^2
Rearranging the equation, we get:
2x^2 = 8
Dividing both sides by 2, we find:
x^2 = 4
Taking the square root of both sides, we get:
x = ±2
So the points of intersection are (2, 4) and (-2, 4).
Step 2: Set up the integral.
To find the volume, we need to integrate the area of each infinitesimally thin disk. The radius of each disk is the distance from the y-axis to the curve. Since we are revolving the region about the y-axis, the radius will be the x-value.
The volume integral is given by:
V = ∫[a,b] π * (radius)^2 dy
where [a,b] are the bounds for the y-values (in this case, from 0 to 4).
Step 3: Express the radius in terms of y.
The x-values are given by the square roots of the y-values because the curves are presented as y = f(x). Thus, we can express the radius as:
radius = |x| = √y
Step 4: Evaluate the integral.
Using the bounds [0,4] and the expression for the radius above, we have:
V = ∫[0,4] π * (√y)^2 dy
Simplifying, we get:
V = ∫[0,4] π * y dy
Evaluating the integral, we get:
V = π * [y^2/2] evaluated from 0 to 4
V = π * (16/2 - 0/2)
V = π * 8
Therefore, the volume of the solid generated by revolving the region about the y-axis is 8π.
To find the volume of the solid generated by revolving the region about the y-axis, we can use the disk method.
First, let's find the points of intersection between the curves y = 8 - x^2 and y = x^2. Setting the two equations equal to each other, we have:
8 - x^2 = x^2
2x^2 = 8
x^2 = 4
x = ±2
So, the region bounded by the curves is between x = -2 and x = 2.
Now, let's consider a small slice of the region at a certain y-value. This slice can be represented as a disk with radius r and thickness Δy.
To find the radius at a certain y-value, we need to express it in terms of y. Looking at the equations of the curves, we can rewrite them as x = √y and x = -√(8 - y).
For the region bounded by -2 ≤ x ≤ 2, we need to consider the positive x-value, so we can use x = √y. Since we are rotating around the y-axis, the radius, r, is equal to x, which is √y.
The thickness, Δy, is a small increase in y, so we can express it as dy.
The volume of the thin disk can be calculated using the formula for the volume of a cylinder: V = π * r^2 * Δy.
Substituting r = √y and Δy = dy, we have:
dV = π * (√y)^2 * dy
dV = π * y * dy
To find the total volume, we need to integrate this expression over the entire region.
∫(from y = 0 to y = 8) π * y * dy
= π * ∫(from y = 0 to y = 8) y dy
Evaluating the integral, we have:
= π * [y^2/2] (from y = 0 to y = 8)
= π * (8^2/2 - 0^2/2)
= π * 32
So, the volume of the solid generated by revolving the region about the y-axis is 32π cubic units.
Therefore, the volume of the solid is 32π cubic units.