면적분과 플럭스

$0$

$-1/2$

$\pi$

By symmetry, the centroid $(\bar x,\bar y,\bar z)$ obeys $\bar x=\bar y=\bar z$. Parametrize the sphere using spherical coordinates.

$$\mathbf{r}(\theta,\varphi) = a\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +a\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +a\cos\varphi\,\hat{\mathbf{k}}$$

Then

$$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta} =-a\sin\varphi\sin\theta\,\hat{\boldsymbol{\imath}}+a\sin\varphi\cos\theta\,\hat{\boldsymbol{\jmath}}\qquad \frac{\partial \mathbf{r}}{\partial \varphi} =a\cos\varphi\cos\theta\,\hat{\boldsymbol{\imath}}+a\cos\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a\sin\varphi\,\hat{\mathbf{k}} \end{aligned}$$

so that

$$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ -a\sin\varphi\sin\theta & a\sin\varphi\cos\theta & 0 \\ a\cos\varphi\cos\theta & a\cos\varphi\sin\theta & -a\sin\varphi \end{matrix}\right] \\ &=-a^2\sin^2\varphi\cos\theta\,\hat{\boldsymbol{\imath}}-a^2\sin^2\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a^2\sin\varphi\cos\varphi\,\hat{\mathbf{k}}\\ \implies \mathrm{d}S&=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi}\right| \,\mathrm{d}\theta\,\mathrm{d}\varphi =a^2\sin\varphi\,\mathrm{d}\theta\,\mathrm{d}\varphi \end{aligned}$$

As the surface area of the part of the sphere in the first octant is $\frac{1}{8}4\pi a^2=\frac{\pi a^2}{2}$

$$\begin{aligned} \bar x=\bar y=\bar z &=\frac{\iint _S z\,\mathrm{d}S}{\iint _S\,\mathrm{d}S} =\frac{1}{\pi a^2/2}\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi \ (a^2\sin\varphi)(\overbrace{a\cos\varphi}^{z}) \\ &=\frac{2a}{\pi}\frac{\pi}{2}\int_0^{\pi/2}\mathrm{d}\varphi \ \sin\varphi\cos\varphi =a\bigg[\frac{1}{2}\sin^2\varphi\bigg]_0^{\pi/2} =\frac{a}{2} \end{aligned}$$

(a) Each (horizontal) constant $z$ cross-section is a circle centred on the $z$-axis. The radius varies linearly from $2$, when $z=0$ to $0$, when $z=3$. So the radius at height $z$ is $\frac{2}{3}(3-z)$ and we can use

$$\begin{aligned} \mathbf{r}(\theta,z) = \frac{2}{3}(3-z)\cos\theta\ \hat{\boldsymbol{\imath}} +\frac{2}{3}(3-z)\sin\theta\ \hat{\boldsymbol{\jmath}} +z\,\hat{\mathbf{k}}\qquad 0\le\theta<2\pi,\quad 0\le z\le 3 \end{aligned}$$

as the parametrization.

(b) By symmetry the centre of mass will lie on the $z$-axis. We are only asked for the $z$-coordinate anyway. The $z$-coordinate of the centre of mass is the weighted average of $z$ over the cone. Since a density has not been specified, we assume that it is a constant. We may take the density to be $1$, so the $z$-coordinate of the centre of mass is $\iint _S z\,\mathrm{d}S/ \iint _S \mathrm{d}S$.

Since

$$\begin{aligned} \tfrac{\partial\mathbf{r}}{\partial\theta} &=\big(-\tfrac{2}{3}(3-z)\sin\theta\,,\, \tfrac{2}{3}(3-z)\cos\theta\,,\, 0 \big) \\ \tfrac{\partial\mathbf{r}}{\partial z} &=\big(-\tfrac{2}{3}\cos\theta\,,\, -\tfrac{2}{3}\sin\theta\,,\, 1 \big) \\ \tfrac{\partial\mathbf{r}}{\partial\theta} \times \tfrac{\partial\mathbf{r}}{\partial z} &=\big(\tfrac{2}{3}(3-z)\cos\theta\,,\, \tfrac{2}{3}(3-z)\sin\theta\,,\, \tfrac{4}{9}(3-z)\big) \end{aligned}$$

the element of surface area for this parametrization is

$$\begin{aligned} \mathrm{d}S &= \big|\tfrac{\partial\mathbf{r}}{\partial\theta}\times \tfrac{\partial\mathbf{r}}{\partial z} \big|\mathrm{d}\theta\mathrm{d}z =\tfrac{2}{3}(3-z)\big|(\cos\theta\,,\,\sin\theta\,,\,\tfrac{2}{3}\big)\big| \mathrm{d}\theta\mathrm{d}z \\ &=\tfrac{2\sqrt{13}}{9}(3-z) \mathrm{d}\theta\mathrm{d}z \end{aligned}$$

So the surface area, $\iint _S \mathrm{d}S$, of the cone is

$$\begin{aligned} \int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z) &=\tfrac{4\sqrt{13}}{9}\pi\int_0^3\mathrm{d}z\ (3-z) =-\tfrac{2\sqrt{13}}{9}\pi(3-z)^2\Big|_0^3 \\ &=2\sqrt{13}\pi \end{aligned}$$

and the $z$-coordinate of the centre of mass is

$$\begin{aligned} \bar z &=\frac{1}{2\sqrt{13}\pi} \int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z) z =\frac{2}{9}\int_0^3\mathrm{d}z\ (3z-z^2) =\frac{2}{9}{\Big[\frac{3z^2}{2}-\frac{z^3}{3}\Big]}_0^3 \\ &=\frac{2}{9}\frac{27}{6} =1 \end{aligned}$$

This is a little less than half way up the cone, which is reasonable since the cone is “bottom heavy”.

Write

$$V=\big\{ \ (x,y,z)\ \big|\ 1\le x^2+y^2+z^2\le 4\ \big\}$$

The boundary of $V$ consists of two parts — the sphere, $S_2$, of radius $2$, centred on the origin, with (outward) normal $\hat{\mathbf{n}}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{\mathbf{r}}{2}$, and the sphere $S_1$ of radius $1$, centred on the origin, with (inward) normal $\hat{\mathbf{n}} = -\mathbf{r}$, So,

$$\begin{aligned} \iint _{\partial V} \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iint _{S_2} \frac{\mathbf{r}}{|\mathbf{r}|}\cdot\frac{\mathbf{r}}{2}\,\mathrm{d}S -\iint _{S_1} \frac{\mathbf{r}}{|\mathbf{r}|}\cdot\mathbf{r}\,\mathrm{d}S \\ &=\iint _{S_2} \mathrm{d}S -\iint _{S_1} \mathrm{d}S \\ &=4\pi(2)^2 -4\pi(1)^2 \\ &=12\pi \end{aligned}$$

First observe that,

• because $(x+y+1)^2\ge 0$, all points on $(x+y+1)^2+z^2=4$ have $|z|\le 2$ and that,
• for $|z_0|\le 2$, the surface $(x+y+1)^2+z^2=4$ intersects the horizontal plane $z=z_0$ on $(x+y+1)^2 = 4-z_0^2$, i.e. on the two lines $x+y=\pm\sqrt{4-z_0^2}-1$, $z=z_0$.
• The line $x+y=\pm\sqrt{4-z_0^2}-1$, $z=z_0$ intersects the first octant if and only if $z_0\ge0$ and $\pm\sqrt{4-z_0^2}-1\ge 0$.
• Thus $x+y=-\sqrt{4-z_0^2}-1$, $z=z_0$ never intersects the first octant and
• $x+y=\sqrt{4-z_0^2}-1$, $z=z_0$ intersects the first octant if and only if $0\le z_0\le \sqrt{3}$.
• When $z_0=0$, the line $x+y=\sqrt{4-z_0^2}-1$, $z=z_0$ is $x+y=1$, $z=0$.
• When $z_0=\sqrt{3}$, the line $x+y=\sqrt{4-z_0^2}-1$, $z=z_0$ is $x+y=0$, $z=2$.
• So as $(x,y,z)$ runs over $\mathcal{S}$, $(x,y)$ runs over the triangle $x\ge0$, $y\ge 0$, $x+y\le 1$.

Let $G(x,y,z)=(x+y+1)^2+z^2$. Then $$\hat{\mathbf{n}}\,\mathrm{d}S=\pm\frac{{ \boldsymbol{\nabla} } G}{{ \boldsymbol{\nabla} } G\cdot\hat{\mathbf{k}}}\ \mathrm{d}x\mathrm{d}y =\pm\frac{2(x+y+1)\hat{\boldsymbol{\imath}}+2(x+y+1)\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}}}{2z}\,\mathrm{d}x\mathrm{d}y$$ For the downward normal, we need the minus sign, so

$$\begin{aligned} \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=-\big[xy\,\hat{\boldsymbol{\imath}}+(z-xy)\,\hat{\boldsymbol{\jmath}}\big]\cdot \Big[\frac{(x+y+1)\hat{\boldsymbol{\imath}}+(x+y+1)\hat{\boldsymbol{\jmath}}+z\hat{\mathbf{k}}}{z}\Big]\,\mathrm{d}x\mathrm{d}y\\ &=-\frac{1}{z}\big[xy(x+y+1)+(z-xy)(x+y+1)\big]\,\mathrm{d}x\mathrm{d}y\\ &=-\frac{1}{z}\big[z(x+y+1)\big]\,\mathrm{d}x\mathrm{d}y\\ &=-(x+y+1)\,\mathrm{d}x\mathrm{d}y \end{aligned}$$

The domain of integration is $x\ge 0,\ y\ge 0,\ x+y\le 1$, so

$$\begin{aligned} \iint _\mathcal{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=-\int_0^1 \mathrm{d}x\int_0^{1-x} \mathrm{d}y \ (x+y+1) =-\int_0^1 \mathrm{d}x\ \Big[(1+x)(1-x)+\frac{1}{2}(1-x)^2\Big]\\ &=-\int_0^1 \mathrm{d}x\ \Big[\frac{3}{2}-x-\frac{1}{2} x^2\Big] =-\Big[\frac{3}{2}-\frac{1}{2}-\frac{1}{6}\Big] =-\frac{5}{6} \end{aligned}$$