Asked by Anonymous
1. Find the volume V obtained by rotating the region bounded by the curves about the given axis.
y = sin(x), y = 0, π/2 ≤ x ≤ π; about the x−axis
2. Find the volume V obtained by rotating the region bounded by the curves about the given axis.
y = 3 sin2(x), y = 0, 0 ≤ x ≤ π; about the x−axis
I am struggling with the setup.
y = sin(x), y = 0, π/2 ≤ x ≤ π; about the x−axis
2. Find the volume V obtained by rotating the region bounded by the curves about the given axis.
y = 3 sin2(x), y = 0, 0 ≤ x ≤ π; about the x−axis
I am struggling with the setup.
Answers
Answered by
Reiny
You want
Volume = π∫ sin^2 x dx from π/2 to π
the hard part is to integrate sin^2 x
start with : cos (2x) = 1 - 2sin^2 x
2sin^2 x = 1 - cos(2x)
sin^2 x = 1/2 - (1/2)cos(2x)
so ∫ sin^2 x dx = (1/2)x - (1/4)sin(2x)
Volume = π∫ sin^2 x dx from π/2 to π
= π[ (1/2)x - (1/4)sin (2x) ] from π/2 to π
= π( π/2 - (1/4)(0) - (π/4 - (1/4)(0) ) )
= π(π/4)
= π^2/4
In the next one you will have to integrate sin^4 x
there are many videos showing you how to do that,
here is "blackpenredpen's " version
https://www.youtube.com/watch?v=SCQdKorKbKM
Volume = π∫ sin^2 x dx from π/2 to π
the hard part is to integrate sin^2 x
start with : cos (2x) = 1 - 2sin^2 x
2sin^2 x = 1 - cos(2x)
sin^2 x = 1/2 - (1/2)cos(2x)
so ∫ sin^2 x dx = (1/2)x - (1/4)sin(2x)
Volume = π∫ sin^2 x dx from π/2 to π
= π[ (1/2)x - (1/4)sin (2x) ] from π/2 to π
= π( π/2 - (1/4)(0) - (π/4 - (1/4)(0) ) )
= π(π/4)
= π^2/4
In the next one you will have to integrate sin^4 x
there are many videos showing you how to do that,
here is "blackpenredpen's " version
https://www.youtube.com/watch?v=SCQdKorKbKM
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