The region R is bounded by the curve y=x +2sinx, x-axis and x=pi, is rotated through 2pi about the x-axis. Find the volume of revolution formedMy answer is (pi^4)/3 + 6(pi^2), but im not sure if it...

The region R is bounded by the curve y=x +2sinx, x-axis and x=pi, is rotated through 2pi about the x-axis. Find the volume of revolution formed

My answer is (pi^4)/3 + 6(pi^2), but im not sure if it is correct

thanks

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the volume of solid of revolutionĀ  formed by rotating the region bounded by the curve `y = x + 2 sin x` , x axis, `x = 0` and `x = pi` , about x axis, such that:

`V = int_0^(pi) pi*f^2(x) dx`

You need to substitute `x + 2 sin x` forĀ  `f(x)` such that:

`V = int_0^(pi) pi*(x + 2 sin x)^2 dx`

Expanding the square yields:

`V = pi*int_0^(pi) (x^2 + 4x*sin x + 4sin^2 x) dx`

Using the property of linarity of integral yields:

`V = pi*int_0^(pi) (x^2) dx + pi*int_0^(pi) 4x*sin x dx + pi*int_0^(pi) 4sin^2 x dx`

`V = pi*(x^3/3)|_0^pi + 4pi*int_0^(pi) x*sin x dx + 4pi*int_0^(pi) sin^2 x dx`

You need to use integration by parts to evaluate the integral `int_0^(pi) x*sin x dx` such that:

`int udv = uv - int vdu`

`u = x => du = dx`

`dv = sin x dx => v = -cos x`

`int_0^(pi) x*sin x dx = -x*cos x_0^pi + int_0^(pi) cos x dx`

`int_0^(pi) x*sin x dx = (-x*cos x + sin x)_0^pi`

You need to use the half angle identity to evaluate the integral `int_0^(pi) sin^2 x dx` such that:

`sin^2 x = (1 - cos 2x)/2`

`int_0^(pi) sin^2 x dx = int_0^(pi)(1 - cos 2x)/2 dx`

`int_0^(pi)(1 - cos 2x)/2 dx = (1/2)int_0^(pi) dx - (1/2)int_0^(pi) cos 2x dx`

`int_0^(pi)(1 - cos 2x)/2 dx = (1/2)(x - (sin 2x)/2)|_0^(pi)`

`int_0^(pi)(1 - cos 2x)/2 dx = (1/2)(pi - (sin 2pi)/2 - 0 + (sin 0)/2)`

`int_0^(pi)(1 - cos 2x)/2 dx = pi/2`

`V = pi*(x^3/3)|_0^pi + 4pi*(-x*cos x + sin x)_0^pi + 4pi*pi/2`

`V = pi*(pi^3/3 - 0^3/3) + 4pi*(sin pi - pi*cos pi - sin 0 + 0*cos0) + 4pi*pi/2`

`V = pi^4/3 + 4pi(0 - pi*(-1) - 0 + 0*1) + 2pi^2`

`V = pi^4/3 + 6pi^2`

Hence, evaluating the volume of solid of revolution, under the given condition, yields that the answer `V = pi^4/3 + 6pi^2` is correct.

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