Question

Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about the​ x-axis. yequals=StartRoot sine Superscript negative 1 Baseline x EndRootsin−1x​, yequals=StartRoot StartFraction pi Over 2 EndFraction EndRoot π 2​, y equals StartRoot StartFraction pi Over 3 EndFraction EndRoot comma and nbspy= π 3, and xequals=0

Answer 1

Volume = 1.535

Explanation:

The region R is bounded by the equations:

y = √sin⁻¹x

y = √(π/2)

y = √(π/3)

x = 0

R is revolved around the x-axis so we will need f(y) for finding out the volume. We need to make x the subject of the equation and then replace it with f(y).

f(x) = √sin⁻¹x

y = √sin⁻¹x

Squaring both sides we get:

y² = sin⁻¹x

x = sin (y²)

f(y) = sin (y²)

Using the Shell Method to find the volume of the solid when R is revolved around the x-axis:

`V = 2\pi \int\limits^a_b {f(y)} \, dy`

The limits a and b are the equations y = √(π/2) and y = √(π/3) which bound the region R. So, a = √(π/2) and b = √(π/3).

V = 2π
`\int\limits^\sqrt{(\pi )/(2)}`
`_\sqrt{(\pi )/(3) }` sin (y²) dy

Integrating sin (y²) dy, we get:

-cos(y²)/2y

So,

V = 2π [-cos(y²)/2y] with limits √(π/2) and √(π/3)

V = 2π [(-cos(√(π/2) ²)/2*√(π/2)] - [(-cos(√(π/3) ²)/2*√(π/3)]

V = 2π [(-cos(π/2)/ 2√(π/2)) - ((-cos(π/3)/ 2√(π/3))]

V = 2π [ 0 - (-0.5/2.0466)]

V = 2π (0.2443)

V = 1.53499 ≅ 1.535