Alphabeta Math
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The solid generated by rotating y=sinx on [0,π] has volume π2/2

Example

Rotate the region

{(x,y):0xπ, 0ysinx}

about the x-axis. The resulting solid is compact and Jordan measurable, and its volume is

V=π22.

The vanishing endpoint radii are included in the solid and require no separate measurability argument.

Facts & Assumptions

Given: The profile f(x)=sinx on [0,π] and its solid of revolution about the x-axis.

[F1]

If ab and f:[a,b][0,) is continuous, then its solid of revolution about the x-axis is compact and Jordan measurable and has volume πabf(x)2dx (The disc formula for the volume of a solid of revolution).

[L1]

sinπ=0, and sinx>0 for every x with 0<x<π; thus π is the first positive zero of sine, in particular π>0 (Pi is the first positive zero of sine).

[L2]

The functions sin and cos are differentiable on R, with (sinx)=cosx and (cosx)=sinx; also sin0=0 and cos0=1 (The derivatives of sine and cosine are cosine and minus sine).

[L3]

For a real-valued function on AR, differentiability at a limit point cA implies continuity there (A function differentiable at c is continuous at c).

[L4]

For every real x, sin2x=(1cos2x)/2 (Double-angle and quadratic power-reduction identities).

[L5]

Every continuous real-valued function on a nondegenerate closed interval is Riemann integrable there (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L8]

If a<b, G:[a,b]R is differentiable at every point of [a,b], and G is integrable, then abG=G(b)G(a) (The second fundamental theorem: if G is differentiable on [a,b] with G=f and f is integrable, then abf=G(b)G(a)).

Verification

technique · direct
1.1

Facts [L1], [L2], and [L3] show that f is continuous and nonnegative on [0,π], so [F1] makes the rotated solid compact and Jordan measurable and gives V=π0πsin2xdx.

F1L1L2L3
1.2

Since π>0 by [L1], the interval [0,π] is nondegenerate. The function G(x)=12sin(2x) is differentiable with G(x)=cos(2x), and this derivative is continuous and therefore integrable on [0,π].

L1L2L3L5L9L10
1.3

On the same nondegenerate interval, the constant function 1 is integrable and has integral π.

L1L7algebra
2.1

By power reduction and linearity, followed by the fundamental theorem applied to step 1.2, 0πsin2xdx=120π1dx120πcos(2x)dx=π/212(G(π)G(0))=π/2, because periodicity and [L2] give G(π)=G(0)=0.

step 1.2step 1.3L2L4L6L8L11algebra
3.1

Substituting step 2.1 into the disc formula of step 1.1 gives V=π(π/2)=π2/2.

step 1.1step 2.1algebra

Remarks

The profile radius vanishes at both endpoints, but [F1] permits nonnegative continuous profiles and explicitly includes zero-radius sections. No division by the profile occurs.

Depends on

Used by

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Sources