Given the area bounded by the curve y = 2sin^2 x and the x - axis between

Question

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)

Answer 1

∫ 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