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Compute the volume of the solid of revolution generated by f(x) = sin x + cos x

by RoRi

Sketch the graph and compute the volume of the solid of revolution generated by:

    \[ f(x) = \sin x + \cos x, \qquad \text{on} \qquad 0 \leq x \leq \pi. \]

The sketch of the ordinate set of f(x) = \sin x + \cos x on [0,\pi] is as follows:

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We then compute the volume of the solid of revolution. We reference the previous exercises here and here for the integrals of \sin^2 x and \cos^2 x.

    \begin{align*} V = \int_0^{\pi} \pi (\sin x + \cos x)^2 \, dx &= \pi \left(\int_0^{\pi} \sin^2 x \, dx + \int_0^{\pi} 2 \sin x \cos x \, dx + \int_0^{\pi} \cos^2 x \, dx \right) \\ &= \pi \left( \frac{\pi}{2} + 0 + \frac{\pi}{2} \right) \\ &= \pi^2. \end{align*}

3 comments

  1. Anonymous says:

    note 

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    which makes it easy to use expansion/contraction of intervals and solve the integral.

  2. richardsull says:

    I think Apostol is guilty of a slight bit of carelessness with this exercise. In the text, he defines ordinate sets only for nonnegative functions. And in the Example in Section 2.12, he also talks about solids of revolution only for nonnegative functions. So I don’t think it makes sense to talk about the ordinate set of this particular function. Moreover, maybe it’s not totally obvious right away that the solid of revolution for that bit of _negative_ area between 3pi/4 and pi ends up giving you a _positive_ volume. It’s not hard to work out that it in fact does, but it’s worth showing.

    Of course you can find the solid of revolution as shown here, or you can find the solid of revolution for the ordinate set of the absolute value of sin x + cos x, and they amount to the same thing (geometrically and algebraically). But I think it’s worth forgiving Apostol for a minor slip.

  3. Hannes says:

    It is also possible, and arguably quicker, to use the identity \sin x + \cos x = \sqrt{2}sin(x+\frac{\pi}{4}, followed by the half-angle formula, and using that the integral of \cos(nx) is 0 over any 2\pi-length interval.

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