Rotate the region enclosed by the given functions around the x-axis. Determine the volume.
y=sqrt(sinx) ; y=0 ; x=0 ;x=pi
limits are 0 to pi
vol = π∫y^2 dx from x=0 to π
= π∫sinx dx from 0 to π
= π [-cosx] from 0 to π
= π (-cosπ - (-cos0) )
= π (1 + 1)
= 2π
To find the volume of the region obtained by rotating the curve around the x-axis, we can use the method of cylindrical shells.
First, let's sketch the region enclosed by the given functions, y = √(sin(x)), y = 0, x = 0, and x = π, to visualize it.
The region is a curve that starts at the origin (0, 0), and its upper boundary is given by y = √(sin(x)). The region extends from x = 0 to x = π, and it is bounded below by the x-axis.
To find the volume, we need to integrate the circumference of each cylindrical shell, multiplied by its height. The radius of each shell is given by the function y = √(sin(x)), and the height is given by the infinitesimal change in x, which is dx.
The formula for the volume using cylindrical shells is:
V = ∫(2π * radius * height) dx
In this case, the radius is √(sin(x)), and the height is dx. Thus, the formula becomes:
V = ∫(2π * √(sin(x)) * dx) from 0 to π
To evaluate this integral, we can use mathematical software or follow these steps:
1. Simplify the expression inside the integral:
V = 2π * ∫(√(sin(x)) * dx) from 0 to π
2. Integrate the function √(sin(x)) with respect to x:
V = 2π * ∫(√(sin(x)) * dx) = 2π * (2/3) * sin^(3/2)(x) from 0 to π
3. Substitute the upper limit π into the expression and then subtract the value obtained when substituting the lower limit 0:
V = 2π * (2/3) * sin^(3/2)(π) - 2π * (2/3) * sin^(3/2)(0)
Simplify further if needed.
By following these steps, you should be able to evaluate the definite integral and obtain the volume of the region.