The volume of the 3-dimensional structure formed by rotating the circle x^2 + (y-5)^2 = 1 around the x-axis can be expressed as V = a\pi^2. What is the value of a?

To find the volume of the 3-dimensional structure formed by rotating the circle around the x-axis, we need to use the method of cylindrical shells. The formula for the volume of a solid of revolution using cylindrical shells is:

V = 2π ∫[a, b] x * h(x) dx

where [a, b] is the interval over which the curve is rotated, x is the distance from the axis of rotation (in this case, the x-axis), and h(x) is the height of the shell at each x value.

In this case, we are rotating the circle x^2 + (y-5)^2 = 1 around the x-axis. To find the height of the shell, we need to find the distance between the top and bottom of the circle at each x value. Since the equation of the circle is y = 5 ± √(1 - x^2), the height of the shell is 2(√(1 - x^2)).

Now we can plug this information into the volume formula:

V = 2π ∫[-1, 1] x * 2(√(1 - x^2)) dx

Simplifying, we have:

V = 4π ∫[0, 1] x√(1 - x^2) dx

To evaluate this integral, we can use a trigonometric substitution. Let x = sin(theta), then dx = cos(theta) d(theta). When x = 0, theta = 0, and when x = 1, theta = π/2.

Substituting and simplifying, we have:

V = 4π ∫[0, π/2] sin(theta)√(1 - sin^2(theta)) cos(theta) d(theta)
= 4π ∫[0, π/2] sin(theta) * cos^2(theta) d(theta)

Now we can use a trigonometric identity to simplify further. The identity is sin(2θ) = 2sin(θ)cos(θ). Rearranging, we have sin(θ) * cos^2(θ) = (1/2)sin(2θ) * cos(θ).

Substituting this back into the integral:

V = 4π ∫[0, π/2] (1/2)sin(2θ) * cos(θ) d(theta)

Now we can integrate:

V = 4π [(1/4)cos^2(θ)]|[0, π/2]
= π/2

So, we have V = (π/2)π^2. Comparing this with the given expression V = aπ^2, we can see that a = π/2.