To find the volume of the solid, we can consider the problem as a collection of infinitely thin slices stacked on top of each other along the x-axis. Each slice is a semicircle with a diameter connecting the graphs of y = sinx and y = -sinx.
First, let's find the equation for the curve that represents the upper half of the semicircle. Since the diameter connects the graphs of y = sinx and y = -sinx, the radius of each semicircle is the difference between these two functions: r(x) = (sinx) - (-sinx) = 2sinx.
To calculate the volume of an individual slice, we need to find its area. The area of a circle is given by A = πr², but since we're dealing with semicircles, we can divide this by 2: A = (1/2)πr².
In this case, the radius of each semicircle is given by r(x) = 2sinx, so we have A(x) = (1/2)π(2sinx)² = 2πsin²x.
Now, to find the volume of the entire solid, we need to integrate the area function A(x) over the interval [0, π]. The volume V can be calculated as follows:
V = ∫[a,b] A(x) dx = ∫[0,π] 2πsin²x dx.
To evaluate this integral, we can use the power-reducing identity for sine: sin²x = (1/2)(1 - cos(2x)). Substituting this into the integral, we have:
V = ∫[0,π] 2π(1/2)(1 - cos(2x)) dx
= π∫[0,π] (1 - cos(2x)) dx.
Now, let's integrate term by term:
V = π∫[0,π] dx - π∫[0,π] cos(2x) dx.
The first integral, ∫(1) dx, is simply x, evaluated from 0 to π, which gives us x = π - 0 = π.
For the second integral, we can use the substitution u = 2x, du = 2dx:
V = π∫[0,π] dx - π∫[0,π] cos(u) (1/2) du
= π[x]∣[0,π] - (π/2)∫[0,2π] cos(u) du
= π[π - 0] - (π/2)[sin(u)]∣[0,2π]
= π² - (π/2)(sin(2π) - sin(0))
= π² - (π/2)(0 - 0)
= π².
Therefore, the volume of the solid enclosed between the graphs of y = sinx and y = -sinx from x = 0 to x = π is π².