To find the volume of the region revolved around the x-axis and y-axis, bounded by the curve y = x^2 and the lines y = 0 and y = 1, we can use the method of cylindrical shells.
First, let's consider revolving the region around the x-axis. We can imagine this as a stack of thin cylindrical shells. Each shell has a height of Δy and a radius of x. To find the volume of each shell, we use the formula V = 2πrh, where r is the radius and h is the height.
To set up the integral, we need to express the radius and height of each shell in terms of y.
The radius of each shell is x, and we can express x in terms of y by taking the square root: x = √y.
The height of each shell is Δy, which represents the vertical thickness of the region at a given value of y.
Now, let's set up the integral to find the volume:
V = ∫[0,1] 2π(√y)(Δy),
where [0,1] represents the interval of y-values over which we are integrating.
To evaluate the integral, we need to express Δy in terms of y.
Since Δy represents the vertical thickness of the region at a given value of y, we can express it as Δy = dy.
Now, we can rewrite the integral:
V = ∫[0,1] 2π(√y)(dy).
To solve the integral, we integrate with respect to y:
V = 2π ∫[0,1] √y dy.
To integrate √y, we use the power rule for integration, which states that ∫x^n dx = (1/(n+1))x^(n+1) + C.
Applying the power rule, we have:
V = 2π [(2/3)y^(3/2)] |[0,1]
= 2π [(2/3)(1^(3/2)) - (2/3)(0^(3/2))]
= 2π [(2/3) - 0]
= 4π/3.
Therefore, the volume of the region revolved around the x-axis is 4π/3.
To find the volume revolved around the y-axis, we need to adjust our approach. Instead of using cylindrical shells, we will use the method of washers.
The washer method involves subtracting the volume of the inner region from the volume of the outer region.
To set up the integral, we need to express the outer radius and inner radius in terms of y.
The outer radius is x, which is √y.
The inner radius is the distance between the curve y = x^2 and the y-axis, which is 0.
The thickness of each washer is Δy, which represents the vertical thickness of the region at a given value of y.
Now, let's set up the integral to find the volume:
V = π ∫[0,1] (√y)^2 - 0^2 dy
= π ∫[0,1] y dy
= π [1/2 y^2] |[0,1]
= π [1/2 (1^2) - 1/2 (0^2)]
= π [1/2 - 0]
= π/2.
Therefore, the volume of the region revolved around the y-axis is π/2.
In summary, the volume of the region revolved around the x-axis is 4π/3, and the volume of the region revolved around the y-axis is π/2.