(a)
z z z z z z z = 0 z=0 z = 0 2 2 2 z = 3 z=3 z = 3 0 0 0 z z z 2 3 ( 3 − z ) \frac{2}{3}(3-z) 3 2 ( 3 − z )
r ( θ , z ) = 2 3 ( 3 − z ) cos θ ı ^ + 2 3 ( 3 − z ) sin θ ȷ ^ + z k ^ 0 ≤ θ < 2 π , 0 ≤ z ≤ 3 \begin{aligned}
\mathbf{r}(\theta,z) = \frac{2}{3}(3-z)\cos\theta\ \hat{\boldsymbol{\imath}}
+\frac{2}{3}(3-z)\sin\theta\ \hat{\boldsymbol{\jmath}} +z\,\hat{\mathbf{k}}\qquad
0\le\theta<2\pi,\quad 0\le z\le 3
\end{aligned} r ( θ , z ) = 3 2 ( 3 − z ) cos θ ^ + 3 2 ( 3 − z ) sin θ ^ + z k ^ 0 ≤ θ < 2 π , 0 ≤ z ≤ 3
이것이 매개변수화이다.
(b)
대칭성에 의해 질량중심은 z z z z z z z z z z z z 1 1 1 z z z ∬ S z d S / ∬ S d S \iint _S z\,\mathrm{d}S/
\iint _S \mathrm{d}S ∬ S z d S / ∬ S d S
∂ r ∂ θ = ( − 2 3 ( 3 − z ) sin θ , 2 3 ( 3 − z ) cos θ , 0 ) ∂ r ∂ z = ( − 2 3 cos θ , − 2 3 sin θ , 1 ) ∂ r ∂ θ × ∂ r ∂ z = ( 2 3 ( 3 − z ) cos θ , 2 3 ( 3 − z ) sin θ , 4 9 ( 3 − z ) ) \begin{aligned}
\tfrac{\partial\mathbf{r}}{\partial\theta}
&=\big(-\tfrac{2}{3}(3-z)\sin\theta\,,\,
\tfrac{2}{3}(3-z)\cos\theta\,,\,
0 \big) \\
\tfrac{\partial\mathbf{r}}{\partial z}
&=\big(-\tfrac{2}{3}\cos\theta\,,\,
-\tfrac{2}{3}\sin\theta\,,\,
1 \big) \\
\tfrac{\partial\mathbf{r}}{\partial\theta} \times \tfrac{\partial\mathbf{r}}{\partial z}
&=\big(\tfrac{2}{3}(3-z)\cos\theta\,,\,
\tfrac{2}{3}(3-z)\sin\theta\,,\,
\tfrac{4}{9}(3-z)\big)
\end{aligned} ∂ θ ∂ r ∂ z ∂ r ∂ θ ∂ r × ∂ z ∂ r = ( − 3 2 ( 3 − z ) sin θ , 3 2 ( 3 − z ) cos θ , 0 ) = ( − 3 2 cos θ , − 3 2 sin θ , 1 ) = ( 3 2 ( 3 − z ) cos θ , 3 2 ( 3 − z ) sin θ , 9 4 ( 3 − z ) )
이므로, 이 매개변수화에 대한 겉넓이 요소는 다음과 같다.
d S = ∣ ∂ r ∂ θ × ∂ r ∂ z ∣ d θ d z = 2 3 ( 3 − z ) ∣ ( cos θ , sin θ , 2 3 ) ∣ d θ d z = 2 13 9 ( 3 − z ) d θ d z \begin{aligned}
\mathrm{d}S &=
\big|\tfrac{\partial\mathbf{r}}{\partial\theta}\times
\tfrac{\partial\mathbf{r}}{\partial z} \big|\mathrm{d}\theta\mathrm{d}z
=\tfrac{2}{3}(3-z)\big|(\cos\theta\,,\,\sin\theta\,,\,\tfrac{2}{3}\big)\big|
\mathrm{d}\theta\mathrm{d}z \\
&=\tfrac{2\sqrt{13}}{9}(3-z) \mathrm{d}\theta\mathrm{d}z
\end{aligned} d S = ∂ θ ∂ r × ∂ z ∂ r d θ d z = 3 2 ( 3 − z ) ( cos θ , sin θ , 3 2 ) d θ d z = 9 2 13 ( 3 − z ) d θ d z
따라서 원뿔의 겉넓이 ∬ S d S \iint _S \mathrm{d}S ∬ S d S
∫ 0 3 d z ∫ 0 2 π d θ 2 13 9 ( 3 − z ) = 4 13 9 π ∫ 0 3 d z ( 3 − z ) = − 2 13 9 π ( 3 − z ) 2 ∣ 0 3 = 2 13 π \begin{aligned}
\int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z)
&=\tfrac{4\sqrt{13}}{9}\pi\int_0^3\mathrm{d}z\ (3-z)
=-\tfrac{2\sqrt{13}}{9}\pi(3-z)^2\Big|_0^3 \\
&=2\sqrt{13}\pi
\end{aligned} ∫ 0 3 d z ∫ 0 2 π d θ 9 2 13 ( 3 − z ) = 9 4 13 π ∫ 0 3 d z ( 3 − z ) = − 9 2 13 π ( 3 − z ) 2 0 3 = 2 13 π
그리고 질량중심의 z z z
z ˉ = 1 2 13 π ∫ 0 3 d z ∫ 0 2 π d θ 2 13 9 ( 3 − z ) z = 2 9 ∫ 0 3 d z ( 3 z − z 2 ) = 2 9 [ 3 z 2 2 − z 3 3 ] 0 3 = 2 9 27 6 = 1 \begin{aligned}
\bar z
&=\frac{1}{2\sqrt{13}\pi}
\int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z) z
=\frac{2}{9}\int_0^3\mathrm{d}z\ (3z-z^2)
=\frac{2}{9}{\Big[\frac{3z^2}{2}-\frac{z^3}{3}\Big]}_0^3 \\
&=\frac{2}{9}\frac{27}{6}
=1
\end{aligned} z ˉ = 2 13 π 1 ∫ 0 3 d z ∫ 0 2 π d θ 9 2 13 ( 3 − z ) z = 9 2 ∫ 0 3 d z ( 3 z − z 2 ) = 9 2 [ 2 3 z 2 − 3 z 3 ] 0 3 = 9 2 6 27 = 1
이는 원뿔의 절반 높이보다 조금 낮은 값인데, 원뿔이 “아래쪽이 무겁기” 때문에 타당하다.