Find the volume of the solid obtained by rotating the region under the graph of the function f(x) = (2)/(x+1) about the x-axis over the interval [0,3]
using discs (washers),
v = ∫[0,3] π(R^2-r^2) dx
where r = 1/2 and R = y = 2/(x+1)
v = π∫[0,3] (4/(x+1)^2 - 1/4) dx
= π (-4/(x+1) - x/4) [0,3]
= π ((-4/4 - 3/4) - (-4/1))
= 9π/4
Using shells,
v = ∫[1/2,2] 2πrh dy
where r=y and h = x = (2/y - 1)
v = 2π∫[1/2,2] y(2/y - 1) dy
= 2π (2y - 1/2 y^2)[1/2,2]
= 9π/4
To find the volume of the solid obtained by rotating the region under the graph of the function f(x) = 2/(x+1) about the x-axis over the interval [0,3], we can use the method of cylindrical shells.
The volume of each cylindrical shell can be calculated using the formula:
V = 2πrhΔx
where r is the radius of the shell at a particular x-value, h is the height of the shell, and Δx is the width of the shell.
In this case, we need to find the expression for r, h, and Δx in terms of x.
The radius of the shell (r) is simply the x-value where the function f(x) intersects the x-axis. In this case, the function intersects the x-axis at f(x) = 0, so we have:
0 = 2/(x+1)
Cross-multiplying, we get:
0 = 2
x + 1 = 0
Solving this equation, we find that x = -1. However, this x-value is outside the interval [0,3]. Therefore, we don't have any shells with a radius of 0, and we can ignore this case.
The height of the shell (h) is given by the function f(x) = 2/(x+1). Therefore, h = f(x) = 2/(x+1).
The width of the shell (Δx) is simply the change in x-values within the interval [0,3]. Therefore, Δx = 3 - 0 = 3.
Now, we can calculate the volume of each cylindrical shell using the formula V = 2πrhΔx:
V = 2π(2/(x+1))(3)
To find the total volume of the solid, we need to integrate this expression over the interval [0,3]:
Total volume = ∫[0,3] 2π(2/(x+1))(3) dx
Evaluating this integral will give us the volume of the solid obtained by rotating the region under the graph of the function f(x) = 2/(x+1) about the x-axis over the interval [0,3].