Given the area bounded by the curve y = 2sin^2 x and the x - axis between
x = 0 and x = pi.
Calculate:
.1 the area bounded. (4)
.2 the volume of the solid of revolution if this area rotates
around the x - axis. (6)
∫ sin² ( x ) dx
Write sin² ( x ) as:
1 / 2 - 1 / 2 cos ( 2 x)
∫ sin² ( x ) dx = ∫ 1 / 2 dx - 1 / 2 ∫ cos ( 2 x ) dx
∫ sin² ( x ) dx = 1 / 2 x - 1 / 2 ∫ cos ( 2 x ) dx
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∫ cos ( 2 x ) dx
Substitution:
u = 2 x
du = 2 dx
dx = du / 2
∫ cos ( 2 x ) dx = ∫ cos ( u ) du / 2 = 1 / 2 ∫ cos ( u ) du = 1 / 2 sin ( u ) + C
∫ cos ( 2 x ) dx = 1 / 2 sin ( 2 x ) + C
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∫ sin² ( x ) dx = 1 / 2 x - 1 / 2 [ 1 / 2 sin ( 2 x ) ] + C
∫ sin² ( x ) dx = 1 / 2 x - 1 / 4 sin ( 2 x ) + C
∫ sin² ( x ) dx = 2 / 4 x - 1 / 4 sin ( 2 x ) + C
∫ sin² ( x ) dx = [ 2 x - sin ( 2 x ) ] / 4 + C
1.
0
∫ sin² ( x ) dx = [ 2 ∙ 2 π - sin ( 2 π ) ] / 4 - [ 2 ∙ 0 - sin ( 0 ) ] / 4 =
2π
[ 4 π - 0 ] / 4 - ( 0 - 0 ) / 4 = 4 π / 4 - 0 = π
The area bounded_
0
∫ sin² ( x ) dx = π
2π
2.
The volume of the solid formed by revolving the region bounded by the curve y=f(x) and the x−axis between x = a and x = b about the x- axis is given by:
V =
b
∫ π [ f(x)² ] dx
a
In this case:
2π
∫ π [ sin( x²)² ] dx = π sin⁴ ( x ) dx
0
2π
∫ π sin⁴ ( x ) dx
0
You solve that.
∫ sin⁴ ( x ) dx = [ sin ( 4 x ) - 8 sin ( 2 x ) +12 x ] / 32 + C
V =
2π
∫ π sin⁴ ( x ) dx = 3 π² / 4
0
1. To find the area bounded by the curve y = 2sin^2x and the x-axis between x = 0 and x = pi, we can integrate the absolute value of the curve.
The curve y = 2sin^2x extends from x = 0 to x = pi. From the graph of y = sin^2x, we can see that the curve is always positive between these limits. Therefore, we don't need to take the absolute value.
The integral of y = 2sin^2x with respect to x between x = 0 and x = pi can be calculated as follows:
Integral[0 to pi] (2sin^2x) dx
= 2 * Integral[0 to pi] (sin^2x) dx
To evaluate this integral, we can use the identity sin^2x = (1 - cos(2x)) / 2:
= 2 * Integral[0 to pi] ((1 - cos(2x)) / 2) dx
= (1 / 2) * Integral[0 to pi] (1 - cos(2x)) dx
Integrating both terms separately:
= (1 / 2) * [x - (1 / 2) * sin(2x)] evaluated from x = 0 to x = pi
= (1 / 2) * [(pi - (1 / 2) * sin(2pi)) - (0 - (1 / 2) * sin(0))]
= (1 / 2) * [(pi - 0) - (0 - 0)]
= (1 / 2) * pi
= pi / 2
Therefore, the area bounded by the curve y = 2sin^2x and the x-axis between x = 0 and x = pi is pi / 2.
2. To find the volume of the solid of revolution when this area rotates around the x-axis, we can use the formula for the volume of a solid of revolution.
The formula for the volume of a solid of revolution when rotating around the x-axis is:
Volume = Integral[a to b] (pi * (f(x))^2) dx
In this case, the function f(x) is given by f(x) = 2sin^2x and the limits of integration are a = 0 and b = pi.
Therefore, the volume is:
Volume = Integral[0 to pi] (pi * (2sin^2x)^2) dx
= pi * Integral[0 to pi] (4sin^4x) dx
Using the trigonometric identity sin^2x = (1 - cos(2x)) / 2, we can rewrite this as:
= pi * Integral[0 to pi] (4 * (1 - cos(2x))^2 / 4) dx
Simplifying:
= pi * Integral[0 to pi] (1 - cos(2x))^2 dx
Expanding (1 - cos(2x))^2:
= pi * Integral[0 to pi] (1 - 2cos(2x) + cos^2(2x)) dx
Integrating each term separately:
= pi * [x - sin(2x) / 2 + (1 / 4) * Integral[0 to pi] (1 + cos(4x)) dx]
= pi * [x - sin(2x) / 2 + (1 / 4) * (x + (1 / 4) * sin(4x)) evaluated from x = 0 to x = pi]
= pi * [(pi - sin(2pi) / 2 + (1 / 4) * (pi + (1 / 4) * sin(4pi))) - (0 - sin(0) / 2 + (1 / 4) * (0 + (1 / 4) * sin(0)))]
= pi * [(pi - 0 / 2 + (1 / 4) * (pi + 0)) - (0 - 0 / 2 + (1 / 4) * (0 + 0))]
= pi * [(pi / 2 + (1 / 4) * pi) - 0]
= pi * (pi / 2 + (1 / 4) * pi)
= pi^2 / 2 + pi^2 / 4
= (3 * pi^2) / 4
Therefore, the volume of the solid of revolution when the area bounded by the curve y = 2sin^2x and the x-axis between x = 0 and x = pi rotates around the x-axis is (3 * pi^2) / 4.
To find the area bounded by the curve y = 2sin^2(x) and the x-axis between x = 0 and x = pi, we need to integrate the function with respect to x.
1. The area bounded by a curve y = f(x) and the x-axis between x = a and x = b can be found using the definite integral:
Area = ∫[a, b] f(x) dx
In this case, the function is y = 2sin^2(x), and the limits of integration are a = 0 and b = pi. Thus, the area can be calculated as:
Area = ∫[0, pi] 2sin^2(x) dx
To solve this integral, we can make use of the double-angle identity for sine:
sin^2(x) = (1 - cos(2x)) / 2
Now, the integral becomes:
Area = ∫[0, pi] 2 * (1 - cos(2x)) / 2 dx
Simplifying further:
Area = ∫[0, pi] (1 - cos(2x)) dx
The integral of 1 with respect to x is x, and the integral of cos(2x) with respect to x is sin(2x)/2. So:
Area = [x - (sin(2x)/2)] evaluated from 0 to pi
Plugging in the limits of integration:
Area = [pi - (sin(2pi)/2)] - [0 - (sin(2*0)/2)]
= pi
Therefore, the area bounded by the curve y = 2sin^2(x) and the x-axis between x = 0 and x = pi is 4 square units.
2. To calculate the volume of the solid of revolution when this area is rotated around the x-axis, we can use the method of cylindrical shells.
The volume of a solid of revolution can be found using the formula:
Volume = ∫[a, b] 2πx * f(x) dx
In this case, the function is y = 2sin^2(x), and the limits of integration are still a = 0 and b = pi. Thus, the volume can be calculated as:
Volume = ∫[0, pi] 2πx * 2sin^2(x) dx
Simplifying further:
Volume = 4π ∫[0, pi] x * sin^2(x) dx
To evaluate this integral, we need to use integration by parts. Let's assign:
u = x --> du = dx
dv = sin^2(x) dx --> v = ∫ sin^2(x) dx = x/2 - sin(2x)/4
Using integration by parts:
Volume = 4π [ x * (x/2 - sin(2x)/4) - ∫(x/2 - sin(2x)/4) dx ]
= 4π [ (x^2)/2 - (x * sin(2x))/4 - ∫(x/2 - sin(2x)/4) dx ]
Simplifying further:
Volume = 4π [ (x^2)/2 - (x * sin(2x))/4 - (x^2)/4 + (sin(2x)/8) ] evaluated from 0 to pi
Plugging in the limits of integration:
Volume = 4π [ ((pi^2)/2 - (pi * sin(2pi))/4 - (pi^2)/4 + (sin(2pi)/8)) - ((0^2)/2 - (0 * sin(2*0))/4 - (0^2)/4 + (sin(2*0)/8)) ]
= 4π [ (pi^2)/2 - (pi * 0)/4 - (pi^2)/4 + 0/8 - 0 ]
= 4π * (pi^2)/4
= π * (pi^2)
Therefore, the volume of the solid of revolution when the bounded area is rotated around the x-axis is 6π cubic units.