Question

Max-Min The arch \(y=\sin x, 0 \leq x \leq \pi,\) is revolved about the ine \(y=c, 0 \leq c \leq 1,\) to generate the solid in the figure. (a) Find the value of c that minimizes the volume of the solid. What is the minimum volume? (b) What value of \(c\) in \([0,1]\) maximizes the volume of the solid? (c) Writing to Learn Graph the solid's volume as a function of \(c,\) first for \(0 \leq c \leq 1\) and then on a larger domain. What happens to the volume of the solid as \(c\) moves away from \([0,1] ?\) Does this make sense physically? Give reasons for your answers.

Short answer

The volume of the solid is minimized at \(c = 0\) with volume \(\pi \beta\), and maximized at \(c = 1\) with volume \(-\pi \gamma\). The volume does not depend on \(c\) in \([0,1]\), and changes beyond this interval due to positive and negative regions as well as extending beyond the original arch.

Step-by-step solution

Determine Volume Equation

The volume of the solid is given by the formula \(\int_{a}^{b} \pi[f(x)]^2 dx - \int_{a}^{b} \pi[g(x)]^2 dx\), where \(f(x) = \sin x\) and \(g(x) = c\). Therefore, the volume is given by \(\int_{0}^{\pi} \pi[(\sin x)^2 - c^2] dx = \pi \int_{0}^{\pi} (\sin^2 x - c^2) dx\).

Differentiate Volume Equation

Taking the derivative of the volume equation gives \(\frac{dV}{dc} = 2\pi c \int_{0}^{\pi}dx - 2c\pi\int_{0}^{\pi}dx\). The integral of \(dx\) over \([0, \pi]\) is just \(\pi\) so the derivative simplifies to \(\frac{dV}{dc} = 2\pi^2c - 2\pi^2c = 0\). Since the derivative is zero for all \(c\), the volume is minimized and maximized when \(c\) is at the endpoints of its domain.

Find Minimum Volume

Substitute \(c = 0\) into volume formula, the minimum volume is \(\pi \int_{0}^{\pi} \sin^2 x dx\). To evaluate this integral, use power-reduction identity: \(\sin^2 x = \frac{1 - \cos 2x}{2}\). The result of the integral is \(\pi \beta\), for some known constant of integration \(\beta\).

Find Maximum Volume

Substitute \(c = 1\) into volume formula, the maximum volume is \(\pi \int_{0}^{\pi} (\sin^2 x - 1) dx\). Again, use power-reduction identity to evaluate the integral, leading to maximum volume \(-\pi \gamma\), for some constant of integration \(\gamma\).

Analyze Volume as a Function of c

The volume of the solid does not depend on \(c\) in \([0,1]\) as \(\frac{dV}{dc} = 0\). However, for \(c < 0\) and \(c > 1\), the solid will include negative regions and extend beyond the original arch respectively, leading to volume changes. This is physically reasonable since changing the axis of rotation will modify the resulting solid.