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​ y-axis. yequals startroot 169 minus 13 x squared endroot​, yequals ​0, and xequals ​0, in the first quadrant set up the integral that gives the volume of the solid using the shell method. use increasing limits of integration. select the correct choice below and fill in the answer boxes to complete your choice.


Volume = 1.535

Step-by-step 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{\frac{\pi }{2}}_\sqrt{\frac{\pi }{3} }   sin (y²) dy

Integrating sin (y²) dy, we get:



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

Do you know the answer?

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