선적분

We use the centre of mass formulae $\bar x = \dfrac{\int_C x\rho\,\mathrm{d}s}{\int_C \rho\,\mathrm{d}s}$, etc. To make the working clearer, we'll break these calculations into several steps.

$$\begin{aligned} x(t)&= t+ \frac12 t^2 & x'(t)&=1+t\\ y(t)&= t- \frac12 t^2 & y'(t)&=1-t\\ z(t)&=\frac43 t^{3/2} & z'(t)&=2\sqrt{t} \end{aligned}$$

$$\begin{aligned} \sqrt{x'(t)^2+y'(t)^2+z'(t)^2}&=\sqrt{1+2t+t^2+1-2t+t^2+4t}=\sqrt{2(t^2+2t+1)}= \sqrt{2}(t+1)\\ \rho(x(t),y(t),z(t))&= \frac{x(t)+y(t)}{2}={\frac{(t+t^2/2 )+( t - t^2/2)}{2}}= t \end{aligned}$$

$$\begin{aligned} \int_C {red}\rho \,\mathrm{d}s&=\int_0^4 t\, \sqrt{2}(t+1)\,\mathrm{d}t =\frac{2^3\cdot 11\sqrt{2}}{3}\\ \int_C x {red}\rho \,\mathrm{d}s&=\int_0^4 \left(t+\frac12t^2\right) t\, \sqrt{2}(t+1)\,\mathrm{d}t =\sqrt{2}\int_0^{2^2} \left(\frac{t^4}{2}+\frac{3}{2}t^3 +t^2\right)\ \mathrm{d}t\\ &=\sqrt{2}\left(\frac{2^9}{5} +3(2^5)+\frac{2^6}{3}\right) =\frac{2^5\cdot103\sqrt 2}{15}\\ \int_C y {red}\rho \,\mathrm{d}s&=\int_0^4 \left(t-\frac12t^2\right) t\, \sqrt{2}(t+1)\,\mathrm{d}t=\sqrt2\int_0^{2^2}\left(-\frac{t^4}{2}+\frac{t^3}{2}+t^2 \right)\mathrm{d}t\\ &=\sqrt{2}\left(-\frac{2^9}{5}+{2^5}+\frac{2^6}{3} \right) =-\frac{2^5\cdot 23\sqrt 2}{15}\\ \int_C z {red} \rho\, \mathrm{d}s&=\int_0^4 \left(\frac{4}{3}t^{3/2}\right) t\, \sqrt{2}(t+1)\,\mathrm{d}t=\frac{4\sqrt2}{3}\int_0^{2^2} \left( t^{7/2}+t^{5/2}\right)\mathrm{d}t\\ &=\frac{4\sqrt2}{3}\left( \frac{2^{10}}{9}+\frac{2^8}{7}\right)=\frac{2^{10}\cdot37\sqrt{2}}{7\cdot 3^3} \\ \overline{x}&=\frac{\int x \rho\,\mathrm{d}s}{\int \rho\,\mathrm{d}s}=\frac{\frac{2^5\cdot103\sqrt 2}{15}}{\frac{2^3\cdot 11\sqrt{2}}{3}}=\frac{412}{55}\approx 7.5\\ \overline{y}&=\frac{\int y \rho\,\mathrm{d}s}{\int \rho\,\mathrm{d}s}=\frac{-\frac{2^5\cdot 23\sqrt 2}{15}}{\frac{2^3\cdot 11\sqrt{2}}{3}}=-\frac{92}{55}\approx-1.7\\ \overline{z}&=\frac{\int z \rho\,\mathrm{d}s}{\int \rho\,\mathrm{d}s}=\frac{\frac{2^{10}\cdot37\sqrt{2}}{7\cdot 3^3}}{\frac{2^3\cdot 11\sqrt{2}}{3}}=\frac{4736}{693}\approx 6.8 \end{aligned}$$

After these long calculations, it's nice to do a sanity check. Using $0 \le t \le 4$, we see our wire takes up space in the following intervals: $0 \le x \le 12$, $-4 \le y \le 1/2$, and $0 \le z \le 32/3$. The coordinates of our centre of mass all fall in these intervals, which doesn't guarantee our answer is correct, but it is a nice sign. If, say $\overline x$ had been negative, or $\overline z$ were greater than 11, we would have known there was something wrong.