밀도·질량과 질량중심

$\frac{2m\pi }{3}$

$\pi {a}^{2}b$

Center of mass: $\left(\frac{3a}{8},\frac{2{a}^{2}}{5}\right),$ volume: $\frac{\pi {a}^{4}}{2}$

(a) Here is a sketch of the specified region, which we shall call $R$.

The top of $R$ has equation $y=T(x)=e^x$, the bottom has equation $y=B(x)=-1$ and $x$ runs from $0$ to $1$. So, using vertical strips, we see that $R$ has area

$$\begin{aligned} A & = \int_0^1 \big[T(x)-B(x)\big]\,\mathrm{d}x = \int_0^1 \big[e^x-(-1)\big]\,\mathrm{d}x = \int_0^1 \big[e^x+1\big]\,\mathrm{d}x = \big[e^x+x\big]_0^1 = e \end{aligned}$$

and

$$\begin{aligned} \bar y &=\frac{1}{2A} \int_0^1 \big[T(x)^2-B(x)^2\big]\,\mathrm{d}x \\ &= \frac{1}{2e}\int_0^1 \big[ e^{2x} - 1\ \big] \,\mathrm{d}x = \frac{1}{2e} \bigg[\frac{e^{2x}}{2} - x \bigg]_0^1 \\ &=\frac{1}{2e} \bigg(\frac{e^2}{2} - 1 - \frac{1}{2} \bigg) = \frac{e}{4} - \frac{3}{4e} \end{aligned}$$

(b) To compute the volume when $R$ is rotated about the line $y=-1$

• we use thin vertical strips of width $\mathrm{d}x$ as in the figure above.

• When we rotate about the line $y=-1$, each strip sweeps out a thin disk

• whose radius is $r=T(x)-B(x)=e^x+1$ and

• whose thickness is $\mathrm{d}x$ and hence

• whose volume is $\pi r^2\,\mathrm{d}x = \pi(e^x+1)^2\,\mathrm{d}x$.

• As our leftmost strip is at $x=0$ and our rightmost strip is at $x=1$, the total volume is

$$\begin{aligned} \pi \int_0^1 (e^x+1)^2\,\mathrm{d}x &=\pi \int_0^1 (e^{2x}+2e^x+1)\,\mathrm{d}x =\pi\bigg[\frac{e^{2x}}{2}+2e^x + x\bigg]_0^1 \\ &=\pi\bigg[\Big(\frac{e^{2}}{2}+2e + 1\Big) -\Big(\frac{1}{2}+2 + 0\Big)\bigg] \\ &=\pi\Big(\frac{e^{2}}{2}+2e -\frac{3}{2}\Big) \end{aligned}$$

(a) The plot shown above (with the curve reflected across the $x$-axis to generate the solid of revolution) shows why this solid of revolution is a reasonable approximation to a baseball bat (if perhaps a bit thin between $x = 3$ and $x = 9$. (b) We use the disk method to find the volume of the solid of revolution. In the usual way, the volume of a typical slice is $$V_{\text{slice}} \approx \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \triangle x$$ and thus using a definite integral to sum the volumes of the slices, we get $$V = \int_0^{30} \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \, dx\,.$$ Using technology to evaluate the integral, we find that $V \approx 52.0666$ cubic inches. (c) Given that the bat has (constant) weight density $0.6$ ounces per cubic inch, it follows that the total weight of the bat is $W \approx 0.6 \cdot 52.0666 = 31.23996$ ounces. (d) If we slice the baseball bat perpendicular to the horizontal axes, we get thin cylindrical disks as cross-sections. At a given $x$-location, the amount of weight concentrated there is approximately the weight density ($0.6$ ounces per cubic inch) times the volume of the slice, which is $V_{text{slice}} \approx \pi f(x)^2$. Thus, we see that the amount of weight concentrated at a location $x$ along the bat is given by $\rho(x) = 0.6 \pi f(x)^2$, which we can therefore view as being the weight density function for how the weight of the baseball bat is distributed from $x = 0$ to $x = 30$. (e) Even though we are dealing with “weight” instead of “mass”, we can compute the center of mass in the usual way since weight differs from mass only by the gravitational constant, and this constant is present in both the numerator and denominator of the center of mass. Thus, using the standard formula, $$\overline{x} = \frac{\int_0^{30} x \rho(x) \ dx}{\int_0^{30} \rho(x) \ dx}$$ so $$\begin{aligned}\overline{x} &= \frac{\int_0^{30} x \cdot 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx} \\ &= \frac{\int_0^{30} x \cdot (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx} \\ &\approx \frac{384.735}{16.5733} = 23.21415\end{aligned}\,,$$ so the center of mass is just over 2/3 along the length of the bat.