The region above the curve
y = sin^-1(x), (0 greater than or equal to y greater than or equal to pi/4), is rotated about the y-axis through a complete revolution to form a solid. Evaluate the volume.
Thanks!
To find the volume of the solid obtained by rotating the region above the curve y = sin^(-1)(x) about the y-axis, we will use the method of cylindrical shells.
First, let's visualize the region in question. The curve y = sin^(-1)(x) is an inverted sine function over the interval 0 ≤ x ≤ 1, and the range of y is π/4 ≤ y ≤ π/2.
Next, we need to find an expression for the height of each cylindrical shell. Since we are rotating the region about the y-axis, the height of each shell will be the difference between the upper and lower y-values of the region.
The lower y-value is π/4 and the upper y-value is π/2, so the height of each shell will be (π/2) - (π/4) = π/4.
Now, let's find an expression for the radius of each cylindrical shell. The radius is the distance from the y-axis to the curve at each y-value. In this case, the curve is y = sin^(-1)(x), so we need to solve for x.
y = sin^(-1)(x)
x = sin(y)
Since we are rotating about the y-axis, the radius will be the value of x at each y-value. So, the radius of each shell is x = sin(y).
Finally, we can integrate to find the volume of the solid:
V = ∫(2πrh) dy
V = ∫(2π(sin(y))(π/4)) dy
= π^2/2 * ∫sin(y) dy
= -π^2/2 * cos(y) + C
To evaluate this integral, we need to find the bounds of integration. Since the region above the curve y = sin^(-1)(x) is bounded by π/4 ≤ y ≤ π/2, the bounds of integration are π/4 and π/2:
V = -π^2/2 * (cos(π/2) - cos(π/4))
= -π^2/2 * (0 - √2/2)
= π^2/4 * √2
Therefore, the volume of the solid obtained by rotating the region above the curve y = sin^(-1)(x) about the y-axis is (π^2/4) * √2.