Find the volume of the solid obtained by rotating the region bounded by the curves y=cos(X), y=0, x=0, x=pi/2 about the line y=-1
Volume=?
using discs of thickness dx,
v = ∫[0,π/2] πr^2 dx
where r = y = cosx
v = ∫[0,π/2] π(cosx)^2 dx = π^2/4
using shells of thickness dy,
v = ∫[0,1] 2πrh dy
where r=y and h = x = arccos(y)
v = ∫[0,1] 2πy*arccos(y) dy = π^2/4
To find the volume of the solid obtained by rotating the region bounded by the curves around the line, we can use the method of cylindrical shells.
First, we need to set up the integral that represents the volume of one cylindrical shell. The height of each shell will be the difference between the y-values of the curves y=cos(x) and y=0, which is cos(x) - 0 = cos(x). The radius of each shell will be the distance from the line y=-1 to the curve y=cos(x), which is cos(x) - (-1) = cos(x) + 1.
The differential volume (dV) of each cylindrical shell can be expressed as dV = 2π(radius)(height)dx. The factor of 2π accounts for the circumference of the shell.
To find the volume of the entire solid, we need to integrate the differential volume over the range of x values from 0 to π/2.
The integral that represents the volume is therefore:
V = ∫(0 to π/2) 2π(cos(x) + 1)(cos(x))dx
To evaluate this integral, we can expand and simplify the expression:
V = ∫(0 to π/2) [2π(cos^2(x) + cos(x))]dx
Next, we can use trigonometric identities to simplify the integrand:
V = ∫(0 to π/2) [2π(1/2 + (1/2)cos(2x) + cos(x))]dx
V = ∫(0 to π/2) [π + πcos(2x) + 2πcos(x)]dx
Now, we can integrate each term separately:
V = [πx + (1/2)πsin(2x) + 2πsin(x)](0 to π/2)
V = [(π/2)π + (1/2)πsin(π) + 2πsin(π/2)] - [0 + (1/2)πsin(0) + 2πsin(0)]
V = [(π^2)/2 + (1/2)πsin(π) + 2π] - [0 + (1/2)πsin(0) + 2πsin(0)]
Finally, we can evaluate the expression:
V = (π^2)/2 + (1/2)πsin(π) + 2π - 0 - 0 - 0
V = (π^2)/2 + (1/2)π + 2π
Therefore, the volume of the solid obtained by rotating the region bounded by the curves y=cos(x), y=0, x=0, x=π/2 about the line y=-1 is:
V = (π^2)/2 + (1/2)π + 2π cubic units.