SOLUTION: Find the volume of the solid obtained by rotating the region bounded by the curves y=cos(x), y=0, x=0, and x=(pi)/2 about the line y=1

Question 967112: Find the volume of the solid obtained by rotating the region bounded by the curves y=cos(x), y=0, x=0, and x=(pi)/2 about the line y=1
Answer by amarjeeth123(570) (Show Source): You can put this solution on YOUR website!
Using the Washer Method:
about y = 1
y = cos(x)
A (x) = π ( outer radius )^2 - π ( inner radius )^2
A (x) = π ( 1 - 0 )^2 - π ( 1 - cos(x) )^2
A (x) = π ( 1 )^2 - π ( 1 - 2cos(x) + cos^2(x) )
A (x) = π ( 1 ) - π ( 1 - 2cos(x) + cos^2(x) )
A (x) = π ( 1 - 1 + 2cos(x) - cos^2(x) )
A (x) = π ( 2cos(x) - cos^2(x) )
π/2
∫ π ( 2cos(x) - cos^2(x) ) dx
0
π/2
∫ π ( 2cos(x) - (1/2) * ( 1 + cos(2x) ) ) dx
0
π/2
∫ π ( 2cos(x) - (1/2) - (1/2) cos(2x) ) dx
0
. . . . . .. . .. . .. . .. . .. . .. . .. . .. . .π/2
π ( 2sin(x) - (1/2) x - (1/4) sin(2x) ) ]
. . .. . .. . .. . .. . .. . .. . .. . .. . .. . .0
π ( 2 * ( sin(π/2) - sin(0) ) - (1/2) ( π/2 - 0 ) - (1/4) * ( sin(2 * π/2) - sin(2 * 0) ) ) ]
π ( 2 * ( 1 - 0 ) - (1/2) (π/2) - (1/4) * ( sin(π) - sin(0) ) ) ]
π ( 2 * 1 - (π/4) - (1/4) * ( 0 - 0)
π ( 2 - (π/4) - 0 )
π ( 2 - (π/4) )
2π - (π^2/4)

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