010. Shell Integration

Given f(x)=πx2sinπx2f(x) = \pi x^2 \sin \pi x^2, calculate the volume VV obtained by rotating the area between f(x)(0x1)f(x) (0 \leq x \leq 1) and xx-axis around yy-axis.


After drawing f(x)f(x), it is found that f(x)f(x) can have two intersections with a line that is perpendicular to the rotation axis. So, it is hard to apply the basic volume calculation method. In this case, shell integration is much more useful. As shown below, the volume ΔV\Delta V obtained by rotating the area between f(x)(txt+Δt)f(x) (t \leq x \leq t + \Delta t) and xx-axis around yy-axis can be approximated as follows.

010-2 ΔVπ{(t+Δt)2t2}f(t)=π{2tΔt+(Δt)2}f(t)    ΔVΔt=π(2t+Δt)f(t)    limΔt0ΔVΔt=dVdt=2πtf(t)\begin{aligned} \Delta V &\approx \pi \{ (t + \Delta t)^2 - t^2 \} f(t) = \pi \{ 2t\Delta t + (\Delta t)^2 \} f(t) \\\\ \implies \frac{\Delta V}{\Delta t} &= \pi (2t + \Delta t) f(t) \implies \lim_{\Delta t \to 0} \frac{\Delta V}{\Delta t} = \frac{dV}{dt} = 2\pi t f(t) \end{aligned}

Note that ΔV\Delta V can be also approximated as π{(t+Δt)2t2}f(t+Δt)\pi \{ (t + \Delta t)^2 - t^2 \} f(t + \Delta t). In this case, however, the same result is induced. Therefore, VV is V=012πtf(t)dt=012πxf(x)dx=012π2x3sinπx2dx=0πusinudu(u=πx2,  du=2πxdx)=[ucosu]0π+0πcosudu=π\begin{aligned} V &= \int_{0}^{1} 2\pi t f(t) dt = \int_{0}^{1} 2\pi x f(x) dx = \int_{0}^{1} 2\pi^2 x^3 \sin \pi x^2 dx \\\\ &= \int_{0}^{\pi} u \sin u du \quad (u = \pi x^2, \; du = 2\pi x dx) \\\\ &= \left[ -u \cos u \right]_{0}^{\pi} + \int_{0}^{\pi} \cos u du = \pi \end{aligned}

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