Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line: y=x, y=x^(1/2); about x=2
To find the volume of the solid obtained by rotating the region bounded by the curves y = x and y = x^(1/2) about the line x = 2, we can use the method of cylindrical shells.
Step 1: Determine the limits of integration
The region bounded by the curves y = x and y = x^(1/2) can be integrated from x = 0 to x = 1, since these are the points where the curves intersect.
Step 2: Set up the integral
The volume of each cylindrical shell is given by the formula V = 2πrh, where r is the distance from the axis of rotation (x = 2) to the shell, and h is the height of the shell. We need to express r and h in terms of x.
The distance from the axis of rotation to the shell, r, is given by r = x - 2.
The height of each shell, h, is given by h = (x^(1/2)) - x, as it represents the difference between the upper and lower heights of the shell.
Step 3: Integrate to find the volume
To find the volume, we integrate the formula V = 2πrh with respect to x, over the interval [0, 1]:
V = ∫[0,1] 2π((x - 2)(x^(1/2)) - (x - 2)x) dx
Simplifying further:
V = 2π ∫[0,1] (x^(3/2) - 2x^(1/2) - x^2 + 2x) dx
Now, integrate each term separately:
V = 2π (∫[0,1] x^(3/2) dx - ∫[0,1] 2x^(1/2) dx - ∫[0,1] x^2 dx + ∫[0,1] 2x dx)
Using the power rule of integration, we get:
V = 2π ( (2/5)x^(5/2) - (4/3)x^(3/2) - (1/3)x^3 + x^2 ) | [0,1]
Evaluate the expression at the upper limit of integration (x = 1) and subtract the expression evaluated at the lower limit of integration (x = 0):
V = 2π ( (2/5)(1)^(5/2) - (4/3)(1)^(3/2) - (1/3)(1)^3 + (1)^2 ) - (2π ( (2/5)(0)^(5/2) - (4/3)(0)^(3/2) - (1/3)(0)^3 + (0)^2 ))
Simplifying further:
V = 2π ( (2/5) - (4/3) - (1/3) + 1 ) - (2π ( 0 - 0 - 0 + 0 ))
V = 2π ( -11/15 )
Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = x and y = x^(1/2) about the line x = 2 is (-22π)/15 cubic units.