Unit C6: Multiple Integrals单元 C6：多重积分

Evaluate $\displaystyle\iint_{R}(x+2y)\,dA$ over $R=[0,2]\times[1,3]$. Integrate in $y$ first, treating $x$ as constant.在 $R=[0,2]\times[1,3]$ 上求 $\displaystyle\iint_{R}(x+2y)\,dA$。先对 $y$ 积分，把 $x$ 当作常数。

$$\int_{1}^{3}(x+2y)\,dy=\Big[xy+y^{2}\Big]_{y=1}^{y=3}=(3x+9)-(x+1)=2x+8.$$

Now integrate the result in $x$.再把结果对 $x$ 积分。

$$\int_{0}^{2}(2x+8)\,dx=\Big[x^{2}+8x\Big]_{0}^{2}=4+16=20.$$

So the integral equals $20$. Reversing the order gives the same value, a useful check.所以积分等于 $20$。交换积分次序会得到相同的值，这是一个有用的检验。

Evaluate $\displaystyle\iint_{R} x\,e^{y}\,dA$ over $R=[0,3]\times[0,1]$. The integrand separates as a product $g(x)h(y)=x\cdot e^{y}$, and the region is a rectangle, so the double integral factors into a product of one-variable integrals.在 $R=[0,3]\times[0,1]$ 上求 $\displaystyle\iint_{R} x\,e^{y}\,dA$。被积函数可分离为乘积 $g(x)h(y)=x\cdot e^{y}$，且区域为矩形，所以二重积分分解为两个单变量积分之积。

$$\iint_{R} x\,e^{y}\,dA=\left(\int_{0}^{3} x\,dx\right)\!\left(\int_{0}^{1} e^{y}\,dy\right)=\Big[\tfrac{x^{2}}{2}\Big]_{0}^{3}\cdot\Big[e^{y}\Big]_{0}^{1}=\frac{9}{2}\,(e-1).$$

The factoring shortcut works only when both the integrand separates and the limits are constant. The next section shows it fails the moment the region stops being a rectangle.这个分解捷径只在被积函数可分离且积分限为常数时成立。下一节会说明，一旦区域不再是矩形，它就失效了。

The average value of $f$ over $R$ is $\displaystyle f_{\text{avg}}=\frac{1}{\operatorname{area}(R)}\iint_{R} f\,dA$. Take $f(x,y)=x^{2}+y$ on $R=[0,2]\times[0,2]$, area $4$. First the exact integral:$f$ 在 $R$ 上的平均值为 $\displaystyle f_{\text{avg}}=\frac{1}{\operatorname{area}(R)}\iint_{R} f\,dA$。取 $f(x,y)=x^{2}+y$，$R=[0,2]\times[0,2]$，面积为 $4$。先算精确积分：

$$\int_{0}^{2}\!\!\int_{0}^{2}(x^{2}+y)\,dy\,dx=\int_{0}^{2}\Big[x^{2}y+\tfrac{y^{2}}{2}\Big]_{0}^{2}dx=\int_{0}^{2}(2x^{2}+2)\,dx=\Big[\tfrac{2x^{3}}{3}+2x\Big]_{0}^{2}=\frac{16}{3}+4=\frac{28}{3}.$$ $$f_{\text{avg}}=\frac{1}{4}\cdot\frac{28}{3}=\frac{7}{3}\approx 2.33.$$

Compare a midpoint Riemann sum with the four subrectangles cut by $x=1$ and $y=1$, each of area $1$, sampling midpoints $(0.5,0.5),(1.5,0.5),(0.5,1.5),(1.5,1.5)$. The values of $x^{2}+y$ are $0.75,\,2.75,\,1.75,\,3.75$, summing to $9$, so the estimate is $9$ for the integral versus the exact $28/3\approx 9.33$. A coarse grid already lands within $4\%$.将其与中点黎曼和比较：用 $x=1$ 和 $y=1$ 切出四个小矩形，每个面积为 $1$，取中点 $(0.5,0.5),(1.5,0.5),(0.5,1.5),(1.5,1.5)$。$x^{2}+y$ 在这些点的值为 $0.75,\,2.75,\,1.75,\,3.75$，求和得 $9$，于是积分的估计值为 $9$，而精确值为 $28/3\approx 9.33$。如此粗的网格已经落在 $4\%$ 误差以内。

The double integral measures signed volume: where $f<0$ the contribution is negative. Evaluate $\displaystyle\iint_{R}(x-y)\,dA$ over $R=[0,1]\times[0,1]$, where the integrand is positive below the diagonal $y=x$ and negative above it.二重积分度量的是有符号体积：在 $f<0$ 处贡献为负。在 $R=[0,1]\times[0,1]$ 上求 $\displaystyle\iint_{R}(x-y)\,dA$，其中被积函数在对角线 $y=x$ 下方为正、上方为负。

$$\int_{0}^{1}\!\!\int_{0}^{1}(x-y)\,dy\,dx=\int_{0}^{1}\Big[xy-\tfrac{y^{2}}{2}\Big]_{0}^{1}dx=\int_{0}^{1}\Big(x-\tfrac{1}{2}\Big)dx=\Big[\tfrac{x^{2}}{2}-\tfrac{x}{2}\Big]_{0}^{1}=0.$$

The integral is exactly $0$: the positive volume below the diagonal cancels the equal negative volume above it, a consequence of the symmetry $x\leftrightarrow y$ over the square. To get true geometric volume between the surface and the plane you would integrate $|x-y|$ instead, which would give a positive number.积分恰好为 $0$：对角线下方的正体积与上方相等的负体积相互抵消，这是正方形上对称性 $x\leftrightarrow y$ 的结果。要得到曲面与平面之间真正的几何体积，应改为积分 $|x-y|$，那会得到一个正数。

Fix a partition. For each column index $i$, group the Riemann sum by the inner variable. For a fixed $x_{i}^{*}$ the partial sum $\sum_{j} f(x_{i}^{*},y_{j}^{*})\,\Delta y$ is a Riemann sum for the single integral $A(x_{i}^{*})=\int_{c}^{d} f(x_{i}^{*},y)\,dy$. Summing the outer index,固定一个分割。对每个列指标 $i$，按内层变量把黎曼和分组。对固定的 $x_{i}^{*}$，部分和 $\sum_{j} f(x_{i}^{*},y_{j}^{*})\,\Delta y$ 是单变量积分 $A(x_{i}^{*})=\int_{c}^{d} f(x_{i}^{*},y)\,dy$ 的一个黎曼和。再对外层指标求和，

$$\sum_{i}\Big(\sum_{j} f(x_{i}^{*},y_{j}^{*})\,\Delta y\Big)\Delta x\;\longrightarrow\;\int_{a}^{b} A(x)\,dx=\int_{a}^{b}\!\!\int_{c}^{d} f\,dy\,dx.$$

Continuity on the compact rectangle gives uniform continuity, which makes the cross-section function $A(x)$ continuous and forces both limits to the same value. The same argument with the grouping reversed yields the other order.紧矩形上的连续性给出一致连续性，从而使截面函数 $A(x)$ 连续，并迫使两个极限取相同的值。把分组方式反过来，同样的论证给出另一种积分次序。

Where continuity matters.连续性在何处起作用。 The continuity hypothesis is not decorative. Fubini's theorem can fail for functions that are merely bounded but badly discontinuous, or for non-integrable functions where the two iterated integrals exist yet disagree. A classical counterexample on the unit square uses $f(x,y)=\dfrac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}$ near the origin: one order gives $\int_0^1\!\int_0^1 f\,dy\,dx=\tfrac{\pi}{4}$ while the reversed order gives $-\tfrac{\pi}{4}$. The function is not absolutely integrable, so neither iterated value equals a genuine double integral. For the continuous functions of this unit the hypotheses always hold and the two orders agree, which is exactly why reversing the order is a legitimate computational tool.连续性假设并非摆设。对仅仅有界却严重不连续的函数，或对两个累次积分都存在却不相等的不可积函数，富比尼定理可能失效。单位正方形上的一个经典反例在原点附近取 $f(x,y)=\dfrac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}$：一种次序给出 $\int_0^1\!\int_0^1 f\,dy\,dx=\tfrac{\pi}{4}$，而相反次序给出 $-\tfrac{\pi}{4}$。该函数不绝对可积，所以任一累次积分值都不等于真正的二重积分。对本单元中的连续函数，假设总是成立、两种次序一致，这正是交换次序成为合法计算工具的原因。

Evaluate $\displaystyle\iint_{D} 6xy\,dA$ where $D$ is the triangle with vertices $(0,0)$, $(2,0)$, $(2,4)$. As a Type I region, $0\le x\le 2$ and $0\le y\le 2x$.求 $\displaystyle\iint_{D} 6xy\,dA$，其中 $D$ 是顶点为 $(0,0)$、$(2,0)$、$(2,4)$ 的三角形。作为 Type I 区域，$0\le x\le 2$ 且 $0\le y\le 2x$。

$$\int_{0}^{2x} 6xy\,dy=6x\cdot\frac{y^{2}}{2}\Big|_{0}^{2x}=3x(2x)^{2}=12x^{3}.$$ $$\int_{0}^{2} 12x^{3}\,dx=3x^{4}\Big|_{0}^{2}=3\cdot 16=48.$$

The integral equals $48$.积分等于 $48$。

Evaluate $\displaystyle\int_{0}^{1}\!\!\int_{x}^{1} e^{y^{2}}\,dy\,dx$. The inner antiderivative $\int e^{y^{2}}\,dy$ has no elementary form, so reverse the order. The region is $0\le x\le 1$, $x\le y\le 1$, that is the triangle below $y=1$ and above $y=x$. As Type II: $0\le y\le 1$, $0\le x\le y$.求 $\displaystyle\int_{0}^{1}\!\!\int_{x}^{1} e^{y^{2}}\,dy\,dx$。内层原函数 $\int e^{y^{2}}\,dy$ 没有初等形式，所以交换次序。区域为 $0\le x\le 1$，$x\le y\le 1$，即位于 $y=1$ 下方、$y=x$ 上方的三角形。作为 Type II：$0\le y\le 1$，$0\le x\le y$。

$$\int_{0}^{1}\!\!\int_{0}^{y} e^{y^{2}}\,dx\,dy=\int_{0}^{1} y\,e^{y^{2}}\,dy=\frac{1}{2}e^{y^{2}}\Big|_{0}^{1}=\frac{1}{2}(e-1).$$

Compute $\displaystyle\iint_{D}(x+y)\,dA$ where $D$ is bounded by $y=x$ and $y=x^{2}$ between $x=0$ and $x=1$. On $[0,1]$ the line $y=x$ lies above the parabola $y=x^{2}$, so as a Type I region $0\le x\le 1$ and $x^{2}\le y\le x$.计算 $\displaystyle\iint_{D}(x+y)\,dA$，其中 $D$ 在 $x=0$ 与 $x=1$ 之间由 $y=x$ 和 $y=x^{2}$ 围成。在 $[0,1]$ 上直线 $y=x$ 位于抛物线 $y=x^{2}$ 上方，所以作为 Type I 区域，$0\le x\le 1$ 且 $x^{2}\le y\le x$。

$$\int_{x^{2}}^{x}(x+y)\,dy=\Big[xy+\tfrac{y^{2}}{2}\Big]_{x^{2}}^{x}=\Big(x^{2}+\tfrac{x^{2}}{2}\Big)-\Big(x^{3}+\tfrac{x^{4}}{2}\Big)=\tfrac{3}{2}x^{2}-x^{3}-\tfrac{1}{2}x^{4}.$$ $$\int_{0}^{1}\Big(\tfrac{3}{2}x^{2}-x^{3}-\tfrac{1}{2}x^{4}\Big)dx=\Big[\tfrac{1}{2}x^{3}-\tfrac{1}{4}x^{4}-\tfrac{1}{10}x^{5}\Big]_{0}^{1}=\frac{1}{2}-\frac{1}{4}-\frac{1}{10}=\frac{3}{20}.$$

As a Type II check, the inverse curves are $x=y$ (lower bound on $x$, since $y\le x$ means $x\ge y$) and $x=\sqrt{y}$ (upper bound, since $y=x^2$ gives $x=\sqrt y\ge y$ on $[0,1]$), so $0\le y\le 1$, $y\le x\le\sqrt{y}$. Carrying out that order returns the same $\tfrac{3}{20}$, confirming the limit translation.作为 Type II 检验，反解曲线为 $x=y$（$x$ 的下界，因为 $y\le x$ 即 $x\ge y$）和 $x=\sqrt{y}$（上界，因为在 $[0,1]$ 上 $y=x^2$ 给出 $x=\sqrt y\ge y$），所以 $0\le y\le 1$，$y\le x\le\sqrt{y}$。按这一次序计算同样得到 $\tfrac{3}{20}$，验证了积分限的转换。

Compute the area of the region $D$ bounded by $y=x$, $y=2$, and $x=0$ (a triangle with vertices $(0,0)$, $(2,2)$, $(0,2)$), to show how the order choice affects the count of integrals. As Type II it is a single clean integral: for each height $0\le y\le 2$, $x$ runs from the left edge $x=0$ to the line $x=y$.计算由 $y=x$、$y=2$、$x=0$ 围成的区域 $D$（顶点为 $(0,0)$、$(2,2)$、$(0,2)$ 的三角形）的面积，以说明次序选择如何影响积分个数。作为 Type II 它是一个干净的单一积分：对每个高度 $0\le y\le 2$，$x$ 从左边界 $x=0$ 到直线 $x=y$。

$$A=\int_{0}^{2}\!\!\int_{0}^{y} dx\,dy=\int_{0}^{2} y\,dy=2.$$

As Type I the same region is also a single integral, $0\le x\le 2$, $x\le y\le 2$, giving $\int_0^2(2-x)\,dx=4-2=2$, the same area. The lesson is that for this convex triangle either order is one piece, but had the upper boundary been the broken line made of $y=x$ then $y=2$, the Type I description would have stayed simple while a horizontal-slice description of a nonconvex variant could split. Always pick the order with fewer pieces.作为 Type I，同一区域也是单一积分，$0\le x\le 2$，$x\le y\le 2$，给出 $\int_0^2(2-x)\,dx=4-2=2$，面积相同。教训是：对这个凸三角形两种次序都只需一块，但若上边界换成由 $y=x$ 再接 $y=2$ 构成的折线，Type I 的描述仍然简单，而某个非凸变体的水平切片描述就可能要分块。总是选择分块更少的次序。

Suppose a Type I region is $a\le x\le b$, $g_1(x)\le y\le g_2(x)$, and we want the Type II description. The procedure is mechanical once the region is drawn. First find the full $y$-range by reading off the minimum and maximum heights, giving the new outer limits $c\le y\le d$. Then, for a horizontal line at fixed height $y$, the left and right boundaries are obtained by inverting the curves $y=g_1(x)$ and $y=g_2(x)$ to solve for $x$; those inverse branches become $h_1(y)\le x\le h_2(y)$.设一个 Type I 区域为 $a\le x\le b$，$g_1(x)\le y\le g_2(x)$，我们想要它的 Type II 描述。一旦画出区域，步骤就是机械的。首先读出最小和最大高度，得到完整的 $y$ 范围，即新的外层限 $c\le y\le d$。然后，对固定高度 $y$ 处的一条水平线，把曲线 $y=g_1(x)$ 和 $y=g_2(x)$ 反解出 $x$ 得到左右边界；这些反函数分支就成为 $h_1(y)\le x\le h_2(y)$。

The subtlety is that a single Type I region can split into several Type II pieces if a horizontal line crosses the boundary more than twice, and vice versa. For example, the disk $x^2+y^2\le 1$ is $-1\le x\le 1$, $-\sqrt{1-x^2}\le y\le \sqrt{1-x^2}$ as Type I and $-1\le y\le 1$, $-\sqrt{1-y^2}\le x\le \sqrt{1-y^2}$ as Type II, both single pieces because the boundary is convex. A nonconvex region such as an annulus or an L-shape generally requires summing two or three iterated integrals in at least one of the orders. The safe habit is: sketch first, decide the order that gives the fewest pieces and the easiest inner antiderivative, and only then write limits.微妙之处在于：若一条水平线与边界相交超过两次，单个 Type I 区域在 Type II 下可能要分成几块，反之亦然。例如圆盘 $x^2+y^2\le 1$ 作为 Type I 是 $-1\le x\le 1$，$-\sqrt{1-x^2}\le y\le \sqrt{1-x^2}$，作为 Type II 是 $-1\le y\le 1$，$-\sqrt{1-y^2}\le x\le \sqrt{1-y^2}$，两者都只需一块，因为边界是凸的。像圆环或 L 形这样的非凸区域，通常在至少一种次序下需要求和两三个累次积分。稳妥的习惯是：先画图，确定分块最少、内层原函数最简单的次序，然后再写积分限。

Evaluate $\displaystyle\iint_{D} e^{-(x^{2}+y^{2})}\,dA$ where $D$ is the disk $x^{2}+y^{2}\le R^{2}$. In polar, $x^{2}+y^{2}=r^{2}$, $0\le r\le R$, $0\le\theta\le 2\pi$.求 $\displaystyle\iint_{D} e^{-(x^{2}+y^{2})}\,dA$，其中 $D$ 是圆盘 $x^{2}+y^{2}\le R^{2}$。在极坐标下，$x^{2}+y^{2}=r^{2}$，$0\le r\le R$，$0\le\theta\le 2\pi$。

$$\int_{0}^{2\pi}\!\!\int_{0}^{R} e^{-r^{2}}\,r\,dr\,d\theta=\int_{0}^{2\pi} d\theta\cdot\int_{0}^{R} e^{-r^{2}} r\,dr.$$

With $u=r^{2}$, $du=2r\,dr$, the inner integral is $\tfrac{1}{2}(1-e^{-R^{2}})$.令 $u=r^{2}$，$du=2r\,dr$，内层积分为 $\tfrac{1}{2}(1-e^{-R^{2}})$。

$$=2\pi\cdot\frac{1}{2}\big(1-e^{-R^{2}}\big)=\pi\big(1-e^{-R^{2}}\big).$$

Find the volume of the solid below $z=9-x^{2}-y^{2}$ and above the disk $x^{2}+y^{2}\le 9$ in the $xy$-plane. The surface meets the plane $z=0$ exactly on the circle $r=3$, so the volume is $\iint_D (9-x^2-y^2)\,dA$ with $D$ the disk of radius $3$. In polar, $9-x^2-y^2=9-r^2$, and $0\le r\le 3$, $0\le\theta\le 2\pi$.求位于 $z=9-x^{2}-y^{2}$ 下方、$xy$ 平面上圆盘 $x^{2}+y^{2}\le 9$ 上方的立体体积。曲面恰好在圆 $r=3$ 上与平面 $z=0$ 相交，所以体积为 $\iint_D (9-x^2-y^2)\,dA$，其中 $D$ 是半径为 $3$ 的圆盘。在极坐标下，$9-x^2-y^2=9-r^2$，且 $0\le r\le 3$，$0\le\theta\le 2\pi$。

$$V=\int_{0}^{2\pi}\!\!\int_{0}^{3}(9-r^{2})\,r\,dr\,d\theta=\int_{0}^{2\pi}d\theta\int_{0}^{3}(9r-r^{3})\,dr.$$ $$\int_{0}^{3}(9r-r^{3})\,dr=\Big[\tfrac{9r^{2}}{2}-\tfrac{r^{4}}{4}\Big]_{0}^{3}=\frac{81}{2}-\frac{81}{4}=\frac{81}{4}.$$ $$V=2\pi\cdot\frac{81}{4}=\frac{81\pi}{2}.$$

Notice the factor $r$ turned the integrand into a polynomial in $r$ that integrates in one line. The same integral in Cartesian form would demand a trigonometric substitution for the disk boundary.注意因子 $r$ 把被积函数变成了关于 $r$ 的多项式，一行就能积出。同一积分若用直角坐标形式，则需要对圆盘边界做三角代换。

Let $I=\int_{-\infty}^{\infty} e^{-x^{2}}\,dx$. The antiderivative is non-elementary, but the square is a double integral that polar coordinates evaluate. Because the integrand separates over the whole plane,设 $I=\int_{-\infty}^{\infty} e^{-x^{2}}\,dx$。其原函数是非初等的，但它的平方是一个可用极坐标求出的二重积分。由于被积函数在整个平面上可分离，

$$I^{2}=\left(\int_{-\infty}^{\infty} e^{-x^{2}}dx\right)\!\left(\int_{-\infty}^{\infty} e^{-y^{2}}dy\right)=\iint_{\mathbb{R}^{2}} e^{-(x^{2}+y^{2})}\,dA.$$

Switching to polar over the entire plane, $0\le r<\infty$, $0\le\theta\le 2\pi$, and using the disk result of Example 3.1 in the limit $R\to\infty$,在整个平面上转为极坐标，$0\le r<\infty$，$0\le\theta\le 2\pi$，并在 $R\to\infty$ 的极限下使用例 3.1 的圆盘结果，

$$I^{2}=\int_{0}^{2\pi}\!\!\int_{0}^{\infty} e^{-r^{2}}\,r\,dr\,d\theta=2\pi\cdot\frac{1}{2}=\pi.$$

Hence $I=\sqrt{\pi}$. This is the cornerstone normalizing constant of the normal distribution, and the polar trick is the standard way to obtain it.因此 $I=\sqrt{\pi}$。这是正态分布的基石归一化常数，而极坐标技巧正是求得它的标准方法。

Evaluate $\displaystyle\iint_{D}\frac{y}{x^{2}+y^{2}}\,dA$ over the upper half of the annulus $1\le x^{2}+y^{2}\le 4$, $y\ge 0$. In polar, $y=r\sin\theta$ and $x^2+y^2=r^2$, so the integrand becomes $\dfrac{r\sin\theta}{r^{2}}=\dfrac{\sin\theta}{r}$, and the region is $1\le r\le 2$, $0\le\theta\le\pi$.在圆环 $1\le x^{2}+y^{2}\le 4$ 的上半部分（$y\ge 0$）上求 $\displaystyle\iint_{D}\frac{y}{x^{2}+y^{2}}\,dA$。在极坐标下，$y=r\sin\theta$ 且 $x^2+y^2=r^2$，于是被积函数变为 $\dfrac{r\sin\theta}{r^{2}}=\dfrac{\sin\theta}{r}$，区域为 $1\le r\le 2$，$0\le\theta\le\pi$。

$$\iint_{D}\frac{y}{x^{2}+y^{2}}\,dA=\int_{0}^{\pi}\!\!\int_{1}^{2}\frac{\sin\theta}{r}\cdot r\,dr\,d\theta=\int_{0}^{\pi}\sin\theta\,d\theta\int_{1}^{2} dr.$$

The Jacobian $r$ neatly cancels the $1/r$ from the integrand, leaving $\big[-\cos\theta\big]_{0}^{\pi}\cdot(2-1)=2\cdot 1=2$. The cancellation is a recurring benefit: integrands built from $x^2+y^2$ and the polar factor $r$ frequently collapse to something elementary.雅可比行列式 $r$ 恰好抵消被积函数中的 $1/r$，剩下 $\big[-\cos\theta\big]_{0}^{\pi}\cdot(2-1)=2\cdot 1=2$。这种抵消是反复出现的好处：由 $x^2+y^2$ 和极坐标因子 $r$ 构成的被积函数常常会塌缩成初等形式。

A polar grid cell spans angles $[\theta,\theta+d\theta]$ and radii $[r,r+dr]$. Its area is the difference of two circular sectors,一个极坐标网格单元跨越角度 $[\theta,\theta+d\theta]$ 和半径 $[r,r+dr]$。它的面积是两个圆扇形之差，

$$\Delta A=\frac{1}{2}(r+dr)^{2}\,d\theta-\frac{1}{2}r^{2}\,d\theta=\frac{1}{2}\big(2r\,dr+dr^{2}\big)d\theta.$$

Dropping the second-order term $dr^{2}$ leaves $\Delta A=r\,dr\,d\theta$. The same factor appears as the absolute value of the Jacobian $\partial(x,y)/\partial(r,\theta)$, which equals $r$. Explicitly, with $x=r\cos\theta$, $y=r\sin\theta$,舍去二阶项 $dr^{2}$ 后剩下 $\Delta A=r\,dr\,d\theta$。同一因子作为雅可比行列式 $\partial(x,y)/\partial(r,\theta)$ 的绝对值出现，其值为 $r$。具体地，取 $x=r\cos\theta$、$y=r\sin\theta$，

$$\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{vmatrix}=r\cos^{2}\theta+r\sin^{2}\theta=r,$$

which is nonnegative for $r\ge 0$, so $|\,\partial(x,y)/\partial(r,\theta)\,|=r$. The geometric wedge argument and the determinant agree, and Section 7 shows both are instances of the general change-of-variables formula.它在 $r\ge 0$ 时非负，所以 $|\,\partial(x,y)/\partial(r,\theta)\,|=r$。几何楔形论证与行列式一致，第 7 节会说明二者都是一般变量替换公式的特例。

A lamina occupies the region $0\le x\le 1$, $0\le y\le x$ with density $\rho(x,y)=1$. Find its mass and center of mass.一块薄片占据区域 $0\le x\le 1$，$0\le y\le x$，密度 $\rho(x,y)=1$。求它的质量和质心。

$$m=\int_{0}^{1}\!\!\int_{0}^{x} 1\,dy\,dx=\int_{0}^{1} x\,dx=\frac{1}{2}.$$ $$M_{y}=\int_{0}^{1}\!\!\int_{0}^{x} x\,dy\,dx=\int_{0}^{1} x^{2}\,dx=\frac{1}{3},\qquad M_{x}=\int_{0}^{1}\!\!\int_{0}^{x} y\,dy\,dx=\int_{0}^{1}\frac{x^{2}}{2}\,dx=\frac{1}{6}.$$ $$\bar{x}=\frac{M_{y}}{m}=\frac{1/3}{1/2}=\frac{2}{3},\qquad \bar{y}=\frac{M_{x}}{m}=\frac{1/6}{1/2}=\frac{1}{3}.$$

The center of mass is $\left(\tfrac{2}{3},\tfrac{1}{3}\right)$, which lies inside the triangle as expected.质心为 $\left(\tfrac{2}{3},\tfrac{1}{3}\right)$，正如预期地落在三角形内部。

Take the same triangle $0\le x\le 1$, $0\le y\le x$ but now with density $\rho(x,y)=x$, so material is heavier toward the right. Recompute the mass and $\bar x$.取同一三角形 $0\le x\le 1$，$0\le y\le x$，但现在密度为 $\rho(x,y)=x$，所以物质越往右越重。重新计算质量和 $\bar x$。

$$m=\int_{0}^{1}\!\!\int_{0}^{x} x\,dy\,dx=\int_{0}^{1} x\cdot x\,dx=\int_{0}^{1} x^{2}\,dx=\frac{1}{3}.$$ $$M_{y}=\int_{0}^{1}\!\!\int_{0}^{x} x\cdot x\,dy\,dx=\int_{0}^{1} x^{3}\,dx=\frac{1}{4},\qquad \bar{x}=\frac{M_{y}}{m}=\frac{1/4}{1/3}=\frac{3}{4}.$$

The centroid had $\bar x=\tfrac23$; weighting by $x$ pulls the balance point right to $\bar x=\tfrac34$, exactly as physical intuition predicts. Density never cancels from $\bar x$ unless it is constant.形心的 $\bar x=\tfrac23$；用 $x$ 加权后平衡点右移到 $\bar x=\tfrac34$，正与物理直觉相符。除非密度为常数，否则它不会从 $\bar x$ 中约去。

The moment of inertia about the origin (the polar moment) of a lamina is $I_0=\iint_D (x^2+y^2)\,\rho\,dA$. Take the quarter disk $x^2+y^2\le a^2$ in the first quadrant with constant density $\rho=1$. Polar coordinates are natural: $0\le r\le a$, $0\le\theta\le\tfrac{\pi}{2}$, and $x^2+y^2=r^2$.薄片关于原点的转动惯量（极矩）为 $I_0=\iint_D (x^2+y^2)\,\rho\,dA$。取第一象限中的四分之一圆盘 $x^2+y^2\le a^2$，密度为常数 $\rho=1$。极坐标很自然：$0\le r\le a$，$0\le\theta\le\tfrac{\pi}{2}$，且 $x^2+y^2=r^2$。

$$I_0=\int_{0}^{\pi/2}\!\!\int_{0}^{a} r^{2}\cdot r\,dr\,d\theta=\int_{0}^{\pi/2}d\theta\int_{0}^{a} r^{3}\,dr=\frac{\pi}{2}\cdot\frac{a^{4}}{4}=\frac{\pi a^{4}}{8}.$$

The integrand $r^2$ from $x^2+y^2$ combines with the Jacobian $r$ to give $r^3$, the source of the $a^4$ scaling that all rotational-inertia formulas share.由 $x^2+y^2$ 得到的被积函数 $r^2$ 与雅可比行列式 $r$ 结合给出 $r^3$，这正是所有转动惯量公式都具有的 $a^4$ 标度的来源。

When density is constant, the center of mass is the centroid, and symmetry can eliminate an integral entirely. The precise statement: if the region $D$ and the density $\rho$ are symmetric under reflection across a line $\ell$, then the center of mass lies on $\ell$. The proof is a substitution. Suppose $D$ is symmetric across the $y$-axis, so $(x,y)\in D\iff(-x,y)\in D$ and $\rho(-x,y)=\rho(x,y)$. In the moment integral substitute $x\mapsto -x$:当密度为常数时，质心就是形心，而对称性可以彻底消去一个积分。精确表述：若区域 $D$ 与密度 $\rho$ 关于某条直线 $\ell$ 的反射对称，则质心位于 $\ell$ 上。证明是一次代换。设 $D$ 关于 $y$ 轴对称，即 $(x,y)\in D\iff(-x,y)\in D$ 且 $\rho(-x,y)=\rho(x,y)$。在矩积分中代换 $x\mapsto -x$：

$$M_{y}=\iint_{D} x\,\rho(x,y)\,dA=\iint_{D}(-x)\,\rho(-x,y)\,dA=-\iint_{D} x\,\rho(x,y)\,dA=-M_{y},$$

so $M_y=0$ and $\bar x=0$. The integral is forced to vanish by the odd symmetry of the integrand over a symmetric region. For example a disk centered at the origin with uniform density has its centroid at the origin without any computation. Recognizing such symmetry before integrating is one of the most reliable ways to cut multivariable problems down to a single integral, and it is worth checking the region for an axis of symmetry as a first step.于是 $M_y=0$ 且 $\bar x=0$。被积函数在对称区域上的奇对称性迫使该积分为零。例如，密度均匀、以原点为中心的圆盘无需任何计算，其形心就在原点。在积分前识别这种对称性，是把多元问题缩减为单一积分的最可靠方法之一，作为第一步检查区域是否有对称轴很值得。

Find the volume of the solid $E$ in the first octant bounded by the plane $x+y+z=1$. Project onto the $xy$-plane: $D$ is the triangle $x\ge 0$, $y\ge 0$, $x+y\le 1$, and $z$ runs from $0$ to $1-x-y$.求第一卦限中由平面 $x+y+z=1$ 围成的立体 $E$ 的体积。投影到 $xy$ 平面：$D$ 是三角形 $x\ge 0$、$y\ge 0$、$x+y\le 1$，而 $z$ 从 $0$ 到 $1-x-y$。

$$V=\int_{0}^{1}\!\!\int_{0}^{1-x}\!\!\int_{0}^{1-x-y} dz\,dy\,dx=\int_{0}^{1}\!\!\int_{0}^{1-x}(1-x-y)\,dy\,dx.$$

The inner integral, with $a=1-x$, is $\int_{0}^{a}(a-y)\,dy=\tfrac{a^{2}}{2}=\tfrac{(1-x)^{2}}{2}$. Then令 $a=1-x$，内层积分为 $\int_{0}^{a}(a-y)\,dy=\tfrac{a^{2}}{2}=\tfrac{(1-x)^{2}}{2}$。于是

$$V=\int_{0}^{1}\frac{(1-x)^{2}}{2}\,dx=\frac{1}{2}\cdot\frac{(1-x)^{3}}{-3}\Big|_{0}^{1}=\frac{1}{6}.$$

Find the mass of the box $E=[0,1]\times[0,2]\times[0,3]$ with density $\rho(x,y,z)=xyz$. Over a box the triple integral separates and the limits are constant, so it factors into a product of single integrals.求长方体 $E=[0,1]\times[0,2]\times[0,3]$ 在密度 $\rho(x,y,z)=xyz$ 下的质量。在长方体上三重积分可分离且积分限为常数，所以它分解为单变量积分之积。

$$m=\int_{0}^{1}\!\!\int_{0}^{2}\!\!\int_{0}^{3} xyz\,dz\,dy\,dx=\left(\int_{0}^{1}x\,dx\right)\!\left(\int_{0}^{2}y\,dy\right)\!\left(\int_{0}^{3}z\,dz\right).$$ $$=\frac{1}{2}\cdot 2\cdot\frac{9}{2}=\frac{9}{2}.$$

As with double integrals, the product shortcut requires both a box-shaped region and a separable integrand. Over a non-box solid the inner limits depend on the outer variables and the factoring fails.和二重积分一样，乘积捷径既需要长方体形状的区域，也需要可分离的被积函数。在非长方体立体上，内层限依赖于外层变量，分解便失效。

Evaluate $\displaystyle\iiint_{E} z\,dV$ where $E$ is bounded below by the $xy$-plane, above by the paraboloid $z=4-x^{2}-y^{2}$, over the disk $x^{2}+y^{2}\le 4$. Set up the inner $z$-integral first, then finish the planar part in polar.求 $\displaystyle\iiint_{E} z\,dV$，其中 $E$ 在下方以 $xy$ 平面为界、在上方以抛物面 $z=4-x^{2}-y^{2}$ 为界，位于圆盘 $x^{2}+y^{2}\le 4$ 之上。先列出内层 $z$ 积分，再用极坐标完成平面部分。

$$\int_{0}^{4-x^{2}-y^{2}} z\,dz=\frac{1}{2}\big(4-x^{2}-y^{2}\big)^{2}.$$

With $x^2+y^2=r^2$ on the disk $0\le r\le 2$, $0\le\theta\le 2\pi$,在圆盘 $0\le r\le 2$，$0\le\theta\le 2\pi$ 上，$x^2+y^2=r^2$，

$$\iiint_{E} z\,dV=\int_{0}^{2\pi}\!\!\int_{0}^{2}\frac{1}{2}(4-r^{2})^{2}\,r\,dr\,d\theta.$$

Let $u=4-r^2$, $du=-2r\,dr$; as $r:0\to2$, $u:4\to0$. The inner integral is $\tfrac12\int_{0}^{4}u^2\cdot\tfrac12\,du=\tfrac14\cdot\tfrac{64}{3}=\tfrac{16}{3}$.令 $u=4-r^2$，$du=-2r\,dr$；当 $r:0\to2$ 时，$u:4\to0$。内层积分为 $\tfrac12\int_{0}^{4}u^2\cdot\tfrac12\,du=\tfrac14\cdot\tfrac{64}{3}=\tfrac{16}{3}$。

$$\iiint_{E} z\,dV=2\pi\cdot\frac{16}{3}=\frac{32\pi}{3}.$$

A solid can be projected onto any of the three coordinate planes, giving six possible orders of integration ($3$ choices of which variable is innermost, then $2$ for the remaining pair). All six give the same value when $f$ is continuous, by the three-dimensional Fubini theorem, but they differ wildly in convenience. The strategy is to pick the projection on which the shadow region is simplest and the bounding surfaces are single-valued in the chosen inner direction.一个立体可以投影到三个坐标平面中的任一个，给出六种可能的积分次序（先有 $3$ 种选择决定哪个变量在最内层，再有 $2$ 种安排其余一对）。当 $f$ 连续时，由三维富比尼定理，六种次序给出相同的值，但它们在方便程度上天差地别。策略是选择投影阴影区域最简单、且边界曲面在所选内层方向上单值的那个投影。

Concretely, integrate first in the direction along which a line enters and exits the solid through exactly two surfaces. For the tetrahedron of Example 5.1, a vertical line enters at $z=0$ and exits at $z=1-x-y$, two clean surfaces, so $dz$ innermost is ideal. If instead the solid were a half-ball sitting on the $xy$-plane, a vertical line still works, but if the solid were a horizontal cylinder a line in the $z$-direction might pierce the curved wall twice and the flat caps, forcing a split; choosing $dx$ or $dy$ innermost avoids that. The practical recipe is the same as in two dimensions: sketch the solid, identify the entering and exiting surfaces for each candidate inner direction, and choose the direction giving two surfaces and the simplest shadow.具体来说，先沿着一条直线恰好穿过两个曲面进出立体的方向积分。对例 5.1 的四面体，竖直线在 $z=0$ 进入、在 $z=1-x-y$ 离开，两个干净的曲面，所以 $dz$ 在最内层最理想。若立体改为放在 $xy$ 平面上的半球，竖直线仍然适用；但若立体是横放的圆柱，沿 $z$ 方向的直线可能两次穿过曲面侧壁和两个平端面，迫使分块，这时选 $dx$ 或 $dy$ 在最内层就能避免。实用的做法与二维相同：画出立体，对每个候选内层方向辨认进入面和离开面，选出给出两个曲面且阴影最简单的方向。

Compute the volume of the ball $x^{2}+y^{2}+z^{2}\le a^{2}$. In spherical coordinates the region is $0\le\rho\le a$, $0\le\phi\le\pi$, $0\le\theta\le 2\pi$.计算球体 $x^{2}+y^{2}+z^{2}\le a^{2}$ 的体积。在球坐标下区域为 $0\le\rho\le a$，$0\le\phi\le\pi$，$0\le\theta\le 2\pi$。

$$V=\int_{0}^{2\pi}\!\!\int_{0}^{\pi}\!\!\int_{0}^{a}\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta.$$

The factors separate: $\int_{0}^{a}\rho^{2}\,d\rho=\tfrac{a^{3}}{3}$, $\int_{0}^{\pi}\sin\phi\,d\phi=2$, $\int_{0}^{2\pi} d\theta=2\pi$.各因子可分离：$\int_{0}^{a}\rho^{2}\,d\rho=\tfrac{a^{3}}{3}$，$\int_{0}^{\pi}\sin\phi\,d\phi=2$，$\int_{0}^{2\pi} d\theta=2\pi$。

$$V=\frac{a^{3}}{3}\cdot 2\cdot 2\pi=\frac{4}{3}\pi a^{3}.$$

Find the volume of the solid bounded below by the paraboloid $z=x^{2}+y^{2}$ and above by the plane $z=4$. The two surfaces meet where $x^2+y^2=4$, a circle of radius $2$, so the shadow is the disk $0\le r\le 2$, and at fixed $(r,\theta)$ the height runs $r^{2}\le z\le 4$.求在下方以抛物面 $z=x^{2}+y^{2}$ 为界、在上方以平面 $z=4$ 为界的立体体积。两曲面相交于 $x^2+y^2=4$，即半径为 $2$ 的圆，所以阴影是圆盘 $0\le r\le 2$，在固定的 $(r,\theta)$ 处高度从 $r^{2}\le z\le 4$。

$$V=\int_{0}^{2\pi}\!\!\int_{0}^{2}\!\!\int_{r^{2}}^{4} r\,dz\,dr\,d\theta=\int_{0}^{2\pi}\!\!\int_{0}^{2}(4-r^{2})\,r\,dr\,d\theta.$$ $$\int_{0}^{2}(4r-r^{3})\,dr=\Big[2r^{2}-\tfrac{r^{4}}{4}\Big]_{0}^{2}=8-4=4,\qquad V=2\pi\cdot 4=8\pi.$$

Cylindrical coordinates were chosen because the solid is rotationally symmetric about the $z$-axis and the bounding surfaces are simple in $r$ and $z$.选用柱坐标是因为立体关于 $z$ 轴旋转对称，且边界曲面在 $r$ 和 $z$ 中很简单。

Integrate $f=1$ (find the volume) over the region $E$ inside the sphere $\rho\le 1$ and above the cone $\phi\le\tfrac{\pi}{3}$. In spherical the description is a rectangle: $0\le\rho\le 1$, $0\le\phi\le\tfrac{\pi}{3}$, $0\le\theta\le 2\pi$.在球 $\rho\le 1$ 内、圆锥 $\phi\le\tfrac{\pi}{3}$ 上方的区域 $E$ 上积分 $f=1$（即求体积）。在球坐标下描述为一个矩形：$0\le\rho\le 1$，$0\le\phi\le\tfrac{\pi}{3}$，$0\le\theta\le 2\pi$。

$$V=\int_{0}^{2\pi}\!\!\int_{0}^{\pi/3}\!\!\int_{0}^{1}\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta.$$

The factors separate: $\int_{0}^{1}\rho^{2}\,d\rho=\tfrac13$, $\int_{0}^{\pi/3}\sin\phi\,d\phi=\big[-\cos\phi\big]_{0}^{\pi/3}=1-\tfrac12=\tfrac12$, $\int_{0}^{2\pi}d\theta=2\pi$.各因子可分离：$\int_{0}^{1}\rho^{2}\,d\rho=\tfrac13$，$\int_{0}^{\pi/3}\sin\phi\,d\phi=\big[-\cos\phi\big]_{0}^{\pi/3}=1-\tfrac12=\tfrac12$，$\int_{0}^{2\pi}d\theta=2\pi$。

$$V=\frac{1}{3}\cdot\frac{1}{2}\cdot 2\pi=\frac{\pi}{3}.$$

The cone $\phi\le\tfrac{\pi}{3}$ becomes a constant limit on $\phi$, which is exactly why spherical coordinates are the natural choice for cone-and-sphere regions.圆锥 $\phi\le\tfrac{\pi}{3}$ 变成了 $\phi$ 上的一个常数限，这正是球坐标对"圆锥加球面"区域成为自然选择的原因。

A solid cone has base radius $1$ and height $1$, described by $0\le r\le z$, $0\le z\le 1$ in cylindrical coordinates (the surface $r=z$ is the cone $z=\sqrt{x^2+y^2}$). Its density grows with height as $\rho=z$. Find the mass, integrating $r$ first within each disk-slice at height $z$.一个实心圆锥底面半径为 $1$、高为 $1$，在柱坐标下描述为 $0\le r\le z$，$0\le z\le 1$（曲面 $r=z$ 即圆锥 $z=\sqrt{x^2+y^2}$）。其密度随高度增长为 $\rho=z$。求质量，在每个高度 $z$ 的圆盘切片内先对 $r$ 积分。

$$m=\int_{0}^{2\pi}\!\!\int_{0}^{1}\!\!\int_{0}^{z} z\cdot r\,dr\,dz\,d\theta=\int_{0}^{2\pi}\!\!\int_{0}^{1} z\cdot\frac{z^{2}}{2}\,dz\,d\theta.$$ $$=\int_{0}^{2\pi}d\theta\int_{0}^{1}\frac{z^{3}}{2}\,dz=2\pi\cdot\frac{1}{8}=\frac{\pi}{4}.$$

The cylindrical factor $r$ produced the $\tfrac{z^2}{2}$ from the inner integral, and the density $z$ then made the $z$-integral a simple power. Cylindrical coordinates suit the cone because its slope appears as the single linear limit $r=z$.柱坐标因子 $r$ 使内层积分得出 $\tfrac{z^2}{2}$，密度 $z$ 又把 $z$ 积分变成一个简单的幂。柱坐标适合圆锥，因为其斜率表现为单一的线性限 $r=z$。

The Jacobian of the map $(\rho,\phi,\theta)\mapsto(x,y,z)$ is the determinant映射 $(\rho,\phi,\theta)\mapsto(x,y,z)$ 的雅可比行列式为

$$\frac{\partial(x,y,z)}{\partial(\rho,\phi,\theta)}=\begin{vmatrix}\sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta\\ \sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta\\ \cos\phi & -\rho\sin\phi & 0\end{vmatrix}=\rho^{2}\sin\phi.$$

Since $\phi\in[0,\pi]$ makes $\sin\phi\ge 0$, the absolute value is $\rho^{2}\sin\phi$, so $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$. The cylindrical case is the analogous computation with Jacobian $r$.由于 $\phi\in[0,\pi]$ 使 $\sin\phi\ge 0$，其绝对值为 $\rho^{2}\sin\phi$，所以 $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$。柱坐标情形是雅可比行列式为 $r$ 的类似计算。

Expanding the determinant.展开行列式。 Cofactoring along the bottom row, the $\cos\phi$ entry multiplies the $2\times2$ minor $\rho\cos\phi\cos\theta\cdot\rho\sin\phi\cos\theta-(-\rho\sin\phi\sin\theta)\cdot\rho\cos\phi\sin\theta=\rho^2\sin\phi\cos\phi$. The $-\rho\sin\phi$ entry multiplies (with the cofactor sign) the minor $\sin\phi\cos\theta\cdot\rho\sin\phi\cos\theta-(-\rho\sin\phi\sin\theta)\cdot\sin\phi\sin\theta=\rho\sin^2\phi$. Collecting, $\cos\phi\cdot\rho^2\sin\phi\cos\phi+\rho\sin\phi\cdot\rho\sin^2\phi=\rho^2\sin\phi(\cos^2\phi+\sin^2\phi)=\rho^2\sin\phi$. The Pythagorean identity is what collapses the determinant to a single clean factor, mirroring how $\cos^2\theta+\sin^2\theta=1$ produced the bare $r$ in the polar case.沿底行作余子式展开，$\cos\phi$ 项乘以 $2\times2$ 子式 $\rho\cos\phi\cos\theta\cdot\rho\sin\phi\cos\theta-(-\rho\sin\phi\sin\theta)\cdot\rho\cos\phi\sin\theta=\rho^2\sin\phi\cos\phi$。$-\rho\sin\phi$ 项（带余子式符号）乘以子式 $\sin\phi\cos\theta\cdot\rho\sin\phi\cos\theta-(-\rho\sin\phi\sin\theta)\cdot\sin\phi\sin\theta=\rho\sin^2\phi$。合并得 $\cos\phi\cdot\rho^2\sin\phi\cos\phi+\rho\sin\phi\cdot\rho\sin^2\phi=\rho^2\sin\phi(\cos^2\phi+\sin^2\phi)=\rho^2\sin\phi$。正是毕达哥拉斯恒等式把行列式塌缩为一个干净的因子，恰如 $\cos^2\theta+\sin^2\theta=1$ 在极坐标情形中产生了单纯的 $r$。

Evaluate $\displaystyle\iint_{D}(x+y)\,dA$ where $D$ is the parallelogram with edges $u=x-y$, $v=x+y$ running over $0\le v\le 2$, $-1\le u\le 1$. Solve for $x,y$: $x=\tfrac{u+v}{2}$, $y=\tfrac{v-u}{2}$, so $x+y=v$.求 $\displaystyle\iint_{D}(x+y)\,dA$，其中 $D$ 是以 $u=x-y$、$v=x+y$ 为边、范围为 $0\le v\le 2$，$-1\le u\le 1$ 的平行四边形。解出 $x,y$：$x=\tfrac{u+v}{2}$，$y=\tfrac{v-u}{2}$，所以 $x+y=v$。

The Jacobian is雅可比行列式为

$$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}1/2 & 1/2\\ -1/2 & 1/2\end{vmatrix}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$$ $$\iint_{D}(x+y)\,dA=\int_{-1}^{1}\!\!\int_{0}^{2} v\cdot\frac{1}{2}\,dv\,du=\frac{1}{2}\int_{-1}^{1}\!\Big[\frac{v^{2}}{2}\Big]_{0}^{2}\,du=\frac{1}{2}\int_{-1}^{1} 2\,du=2.$$

Evaluate $\displaystyle\iint_{D} xy\,dA$ over the region in the first quadrant bounded by the hyperbolas $xy=1$, $xy=4$ and the lines $y=x$, $y=4x$. The boundaries beg for $u=xy$ and $v=y/x$, so the region becomes the rectangle $1\le u\le 4$, $1\le v\le 4$. Rather than solve for $x,y$, compute the easy forward Jacobian and reciprocate.在第一象限中由双曲线 $xy=1$、$xy=4$ 和直线 $y=x$、$y=4x$ 围成的区域上求 $\displaystyle\iint_{D} xy\,dA$。这些边界自然提示取 $u=xy$ 和 $v=y/x$，于是区域变成矩形 $1\le u\le 4$，$1\le v\le 4$。与其解出 $x,y$，不如计算容易的正向雅可比行列式再取倒数。

$$\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix} y & x\\ -y/x^{2} & 1/x\end{vmatrix}=\frac{y}{x}-\Big(-\frac{x\,y}{x^{2}}\Big)=\frac{y}{x}+\frac{y}{x}=\frac{2y}{x}=2v.$$

Therefore $\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|=\dfrac{1}{|2v|}=\dfrac{1}{2v}$ on the region (where $v>0$), and the integrand is $xy=u$. Hence因此在该区域上（$v>0$）$\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|=\dfrac{1}{|2v|}=\dfrac{1}{2v}$，被积函数为 $xy=u$。于是

$$\iint_{D} xy\,dA=\int_{1}^{4}\!\!\int_{1}^{4} u\cdot\frac{1}{2v}\,du\,dv=\frac{1}{2}\left(\int_{1}^{4}u\,du\right)\!\left(\int_{1}^{4}\frac{dv}{v}\right)=\frac{1}{2}\cdot\frac{15}{2}\cdot\ln 4=\frac{15}{4}\ln 4.$$

Computing $\partial(u,v)/\partial(x,y)$ and taking its reciprocal avoided ever solving the messy system for $x$ and $y$ explicitly.计算 $\partial(u,v)/\partial(x,y)$ 并取其倒数，避免了显式求解关于 $x$ 和 $y$ 的繁琐方程组。

Find the area of the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}\le 1$. Use $x=a\,u$, $y=b\,v$, which maps the unit disk $u^2+v^2\le1$ onto the ellipse. The Jacobian is constant:求椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}\le 1$ 的面积。取 $x=a\,u$、$y=b\,v$，它把单位圆盘 $u^2+v^2\le1$ 映射到椭圆。雅可比行列式为常数：

$$\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix} a & 0\\ 0 & b\end{vmatrix}=ab.$$ $$\operatorname{area}=\iint_{\text{ellipse}} dA=\iint_{u^2+v^2\le1} ab\,du\,dv=ab\cdot(\text{area of unit disk})=ab\cdot\pi=\pi ab.$$

The familiar $\pi ab$ falls out as the unit-disk area scaled by the constant Jacobian $ab$, recovering $\pi r^2$ when $a=b=r$.熟悉的 $\pi ab$ 作为单位圆盘面积乘以常数雅可比行列式 $ab$ 得出，当 $a=b=r$ 时还原为 $\pi r^2$。

The change-of-variables theorem says the local area-stretch factor of a smooth map $T(u,v)=(x(u,v),y(u,v))$ is $\big|\det DT\big|$, the absolute Jacobian. Here is the geometric derivation. Consider a small rectangle in the $uv$-plane with corner $(u_0,v_0)$ and sides $\Delta u$, $\Delta v$ along the axes. Its image under $T$ is, to first order, a parallelogram. The two edges map approximately to the tangent vectors变量替换定理说，光滑映射 $T(u,v)=(x(u,v),y(u,v))$ 的局部面积伸缩因子是 $\big|\det DT\big|$，即雅可比行列式的绝对值。下面是几何推导。考虑 $uv$ 平面中一个小矩形，角点为 $(u_0,v_0)$，沿坐标轴的边长为 $\Delta u$、$\Delta v$。它在 $T$ 下的像到一阶近似是一个平行四边形。两条边近似映射为切向量

$$\mathbf{a}=T(u_0+\Delta u,v_0)-T(u_0,v_0)\approx(x_u,\,y_u)\,\Delta u,\qquad \mathbf{b}=T(u_0,v_0+\Delta v)-T(u_0,v_0)\approx(x_v,\,y_v)\,\Delta v.$$

The area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$ is the magnitude of their cross product, which for planar vectors is the absolute value of the $2\times2$ determinant:由 $\mathbf a$ 与 $\mathbf b$ 张成的平行四边形的面积是它们叉积的大小，对平面向量而言即 $2\times2$ 行列式的绝对值：

$$\text{area}=\big|\,x_u y_v-x_v y_u\,\big|\,\Delta u\,\Delta v=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\Delta u\,\Delta v.$$

Summing these local area distortions over a fine grid and passing to the limit turns the Riemann sum $\sum f\,\Delta A_{xy}$ into $\iint f\,\big|\partial(x,y)/\partial(u,v)\big|\,du\,dv$. The polar factor $r$, the cylindrical factor $r$, and the spherical factor $\rho^2\sin\phi$ are all just this same determinant computed for the respective maps, which is why the entire chapter rests on one theorem. The rigorous proof replaces the first-order approximation with a careful estimate and shows the error vanishes faster than the cell area, but the parallelogram picture captures exactly why the determinant, and nothing else, is the right factor.把这些局部面积畸变在细网格上求和并取极限，就把黎曼和 $\sum f\,\Delta A_{xy}$ 变为 $\iint f\,\big|\partial(x,y)/\partial(u,v)\big|\,du\,dv$。极坐标因子 $r$、柱坐标因子 $r$ 和球坐标因子 $\rho^2\sin\phi$ 都不过是对各自映射计算出的同一个行列式，这正是整章都建立在一个定理之上的原因。严格证明用细致的估计替代一阶近似，并说明误差比单元面积更快地趋于零，但平行四边形的图像恰好揭示了为何行列式（而非别的东西）才是正确的因子。