Sections 1 to 7: double integrals (rectangles, general regions, polar), triple integrals, cylindrical and spherical coordinates, change of variables, and applicationsCALC III第 1 至 7 节：二重积分（矩形区域、一般区域、极坐标），三重积分，柱坐标与球坐标，变量替换及应用CALC III
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PART I · CORE TECHNIQUESComputational fluency · 28 marks计算熟练度 · 28 分
Iterated Integrals and Coordinate Systems累次积分与坐标系
Show all working. Set up limits carefully, state the order of integration, and name the coordinate system used. All answers should be exact.写出完整解题过程。仔细建立积分限，说明积分次序，并注明所用坐标系。所有答案须为精确值。
Q1MEDIUMCOREdouble integral over a rectangle, Fubini's theorem矩形区域上的二重积分，Fubini 定理[8 marks]
Let $R=[0,2]\times[1,3]$. Consider $\displaystyle\iint_{R} f(x,y)\,dA$ for the functions below.设 $R=[0,2]\times[1,3]$。对下列函数考察 $\displaystyle\iint_{R} f(x,y)\,dA$。
(a)Evaluate $\displaystyle\iint_{R}(3x^{2}y + 2y)\,dA$ by integrating first with respect to $x$, then $y$. State Fubini's theorem as you apply it.先对 $x$ 后对 $y$ 积分，计算 $\displaystyle\iint_{R}(3x^{2}y + 2y)\,dA$。应用时请陈述 Fubini 定理。[4]
(b)The function $g(x,y)=xe^{xy}$ is continuous on $R$. Write down both iterated integrals that Fubini guarantees are equal, and evaluate the one whose inner integral is tractable in closed form.函数 $g(x,y)=xe^{xy}$ 在 $R$ 上连续。写出 Fubini 定理保证相等的两个累次积分，并计算其中内层积分可以得到初等闭合形式的那个。[4]
Q2MEDIUMCOREdouble integral over a Type I and Type II region第 I 型与第 II 型区域上的二重积分[8 marks]
Let $D$ be the region bounded by $y=x^{2}$ and $y=2x$.设 $D$ 为由 $y=x^{2}$ 与 $y=2x$ 围成的区域。
(a)Sketch $D$, identify it as a Type I region ($a\le x\le b$, $\varphi_{1}(x)\le y\le\varphi_{2}(x)$), and write down the iterated integral $\displaystyle\iint_{D}(x+y)\,dA$ with correct limits.画出 $D$ 的草图，将其识别为第 I 型区域（$a\le x\le b$，$\varphi_{1}(x)\le y\le\varphi_{2}(x)$），并写出带正确积分限的累次积分 $\displaystyle\iint_{D}(x+y)\,dA$。[3]
(b)Evaluate the integral from part (a).计算 (a) 中的积分。[3]
(c)Re-express the region $D$ as a Type II region ($c\le y\le d$, $\psi_{1}(y)\le x\le\psi_{2}(y)$) and write the corresponding iterated integral (do not evaluate).将区域 $D$ 重新表述为第 II 型区域（$c\le y\le d$，$\psi_{1}(y)\le x\le\psi_{2}(y)$），并写出对应的累次积分（无需计算）。[2]
Q3HARDCOREreversing the order of integration交换积分次序[6 marks]
Consider the iterated integral $\displaystyle\int_{0}^{1}\!\int_{x}^{1} e^{y^{2}}\,dy\,dx$.考察累次积分 $\displaystyle\int_{0}^{1}\!\int_{x}^{1} e^{y^{2}}\,dy\,dx$。
(a)Sketch the region of integration and describe it both as the given Type II strip (inner variable $y$) and as a Type I strip (inner variable $x$).画出积分区域的草图，分别将其描述为所给第 II 型竖条（内层变量为 $y$）和第 I 型横条（内层变量为 $x$）。[3]
(b)Reverse the order of integration and evaluate the resulting integral. (The original order has no elementary antiderivative.)交换积分次序并计算所得积分。（原次序没有初等原函数。）[3]
Q4MEDIUMCOREdouble integral in polar coordinates极坐标下的二重积分[6 marks]
Let $D$ be the disk of radius $3$ centred at the origin.设 $D$ 为以原点为圆心、半径为 $3$ 的圆盘。
(a)Convert $\displaystyle\iint_{D}(x^{2}+y^{2})^{3/2}\,dA$ to polar coordinates, writing the exact limits and the area element $dA=r\,dr\,d\theta$.将 $\displaystyle\iint_{D}(x^{2}+y^{2})^{3/2}\,dA$ 转化为极坐标形式，写出精确的积分限和面积元 $dA=r\,dr\,d\theta$。[3]
(b)Evaluate the polar integral.计算该极坐标积分。[3]
PART II · DEFINITIONS AND PROOFRigorous arguments · 26 marks严格论证 · 26 分
Volume Elements and the Jacobian体积元与 Jacobian 行列式
These items are graded on the logic of the argument. Derive each formula from first principles; do not simply quote a result. State every substitution and verify the sign or sign convention of the Jacobian.此部分按论证逻辑评分。从基本原理推导每个公式，不得直接引用结论。说明每一个换元步骤，并验证 Jacobian 行列式的符号或符号约定。
Q5HARDPROOFderiving the polar area element via the Jacobian通过 Jacobian 推导极坐标面积元[8 marks]
The polar substitution is $x=r\cos\theta$, $y=r\sin\theta$, with $r\ge 0$ and $0\le\theta<2\pi$.极坐标换元为 $x=r\cos\theta$，$y=r\sin\theta$，其中 $r\ge 0$，$0\le\theta<2\pi$。
(a)Compute the Jacobian $\dfrac{\partial(x,y)}{\partial(r,\theta)}$ by evaluating the $2\times 2$ determinant of partial derivatives.通过计算偏导数的 $2\times 2$ 行列式，求 Jacobian $\dfrac{\partial(x,y)}{\partial(r,\theta)}$。[4]
(b)Hence justify why the area element transforms as $dA=r\,dr\,d\theta$, and explain why the factor $r$ is never negative for $r\ge 0$.由此说明面积元为何变换为 $dA=r\,dr\,d\theta$，并解释为何当 $r\ge 0$ 时因子 $r$ 始终非负。[2]
(c)A student writes $dA=dr\,d\theta$ and obtains the wrong answer for the area of a disk. Give the correct area of the disk $r\le a$ from first principles (using the correct $dA$) and state what the student's error produces.某学生写出 $dA=dr\,d\theta$ 并因此得到圆盘面积的错误答案。从基本原理出发（使用正确的 $dA$）给出圆盘 $r\le a$ 的正确面积，并说明该学生的错误会导致什么结果。[2]
Q6MEDIUMPROOFtriple integral over a bounded solid (rectangular coordinates)有界立体上的三重积分（直角坐标）[8 marks]
Let $E$ be the solid tetrahedron bounded by the coordinate planes and the plane $x+y+z=1$ (i.e. $x\ge 0$, $y\ge 0$, $z\ge 0$, $x+y+z\le 1$).设 $E$ 为由坐标平面与平面 $x+y+z=1$ 围成的四面体（即 $x\ge 0$，$y\ge 0$，$z\ge 0$，$x+y+z\le 1$）。
(a)Set up $\displaystyle\iiint_{E} 1\,dV$ as an iterated triple integral in the order $dz\,dy\,dx$, determining all six limits of integration.将 $\displaystyle\iiint_{E} 1\,dV$ 化为积分次序为 $dz\,dy\,dx$ 的累次三重积分，确定全部六个积分限。[4]
(b)Evaluate the integral and confirm that the volume of the tetrahedron equals $\dfrac{1}{6}$.计算该积分，并验证四面体的体积等于 $\dfrac{1}{6}$。[4]
Q7HARDPROOFgeneral change of variables, non-trivial Jacobian一般变量替换，非平凡 Jacobian[10 marks]
Let $D$ be the parallelogram with vertices $(0,0)$, $(2,1)$, $(3,3)$, $(1,2)$. Consider the substitution $u=\frac{1}{5}(2x-y)$, $v=\frac{1}{5}(3y-x)$, which maps $D$ to the unit square $S=[0,1]\times[0,1]$.设 $D$ 为顶点为 $(0,0)$、$(2,1)$、$(3,3)$、$(1,2)$ 的平行四边形。考察换元 $u=\frac{1}{5}(2x-y)$，$v=\frac{1}{5}(3y-x)$，该换元将 $D$ 映射到单位正方形 $S=[0,1]\times[0,1]$。
(a)Invert the substitution to express $x$ and $y$ in terms of $u$ and $v$.对换元求逆，将 $x$ 和 $y$ 表示为 $u$ 和 $v$ 的函数。[3]
(b)Compute the Jacobian $\dfrac{\partial(x,y)}{\partial(u,v)}$ and state its value.计算 Jacobian $\dfrac{\partial(x,y)}{\partial(u,v)}$ 并给出其值。[3]
(c)Hence evaluate $\displaystyle\iint_{D}(2x-y)\,dA$ by changing variables to the unit square.由此通过换元至单位正方形，计算 $\displaystyle\iint_{D}(2x-y)\,dA$。[4]
PART III · APPLICATIONS AND SYNTHESISExtended problems · 28 marks综合应用题 · 28 分
Volume, Mass, and Centroid体积、质量与质心
Set up each problem cleanly. Choose coordinates to exploit symmetry; sketch the solid before writing limits. Carry exact values through all intermediate steps.清晰地建立每道题的积分框架。选择能利用对称性的坐标系；在写出积分限之前先画出立体草图。在所有中间步骤中保留精确值。
Q8HARDAPPLIEDvolume in spherical coordinates, ice-cream cone solid球坐标下的体积，冰淇淋锥形立体[10 marks]
Let $E$ be the solid that lies above the cone $z=\sqrt{x^{2}+y^{2}}$ and inside the sphere $x^{2}+y^{2}+z^{2}=4$.设 $E$ 为位于锥面 $z=\sqrt{x^{2}+y^{2}}$ 上方且在球面 $x^{2}+y^{2}+z^{2}=4$ 内部的立体。
(a)Convert both surfaces to spherical coordinates $(\rho,\phi,\theta)$ where $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$. Determine the range of $\rho$, $\phi$, and $\theta$ that describes $E$.将两个曲面转化为球坐标 $(\rho,\phi,\theta)$，其中 $x=\rho\sin\phi\cos\theta$，$y=\rho\sin\phi\sin\theta$，$z=\rho\cos\phi$。确定描述 $E$ 的 $\rho$、$\phi$ 和 $\theta$ 的范围。[4]
(b)Write the triple integral $\displaystyle\iiint_{E}dV$ in spherical coordinates, using the volume element $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$.用体积元 $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$ 将三重积分 $\displaystyle\iiint_{E}dV$ 写成球坐标形式。[2]
(c)Evaluate the integral to find the volume of $E$.计算该积分，求 $E$ 的体积。[4]
Q9HARDAPPLIEDvolume in cylindrical coordinates柱坐标下的体积[8 marks]
Let $E$ be the solid bounded above by the paraboloid $z=9-x^{2}-y^{2}$ and below by the plane $z=0$ (i.e. inside the cylinder $x^{2}+y^{2}\le 9$, between $z=0$ and $z=9-x^{2}-y^{2}$).设 $E$ 为上方由抛物面 $z=9-x^{2}-y^{2}$ 下方由平面 $z=0$ 所围的立体（即在柱面 $x^{2}+y^{2}\le 9$ 内，夹在 $z=0$ 与 $z=9-x^{2}-y^{2}$ 之间）。
(a)Convert to cylindrical coordinates $(r,\theta,z)$ and state the limits on $r$, $\theta$, and $z$ for the solid $E$. Write the volume element as $dV=r\,dz\,dr\,d\theta$.转化为柱坐标 $(r,\theta,z)$，写出立体 $E$ 中 $r$、$\theta$ 和 $z$ 的积分限。体积元记为 $dV=r\,dz\,dr\,d\theta$。[3]
(b)Evaluate $\displaystyle\iiint_{E}dV$ to find the exact volume of $E$.计算 $\displaystyle\iiint_{E}dV$，求 $E$ 的精确体积。[5]
Q10HARDAPPLIEDmass and centroid of a lamina with variable density变密度薄板的质量与质心[10 marks]
A lamina occupies the region $D$ in the first quadrant bounded by the circle $x^{2}+y^{2}=1$ and the coordinate axes ($x\ge 0$, $y\ge 0$). The density function is $\delta(x,y)=x+y$.一薄板占据第一象限中由圆 $x^{2}+y^{2}=1$ 与坐标轴（$x\ge 0$，$y\ge 0$）围成的区域 $D$，密度函数为 $\delta(x,y)=x+y$。
(a)Convert to polar coordinates and write the iterated integral for the total mass $m=\displaystyle\iint_{D}\delta\,dA$. Evaluate $m$.转化为极坐标，写出总质量 $m=\displaystyle\iint_{D}\delta\,dA$ 的累次积分并计算 $m$。[4]
(b)Set up the moment integrals $M_{y}=\displaystyle\iint_{D}x\,\delta\,dA$ and $M_{x}=\displaystyle\iint_{D}y\,\delta\,dA$. Evaluate both.建立矩积分 $M_{y}=\displaystyle\iint_{D}x\,\delta\,dA$ 和 $M_{x}=\displaystyle\iint_{D}y\,\delta\,dA$，并分别计算。[4]
(c)Find the centroid $(\bar{x},\bar{y})$ of the lamina. Explain briefly why, by symmetry, $\bar{x}=\bar{y}$, and verify that your numerical answer confirms this.求薄板的质心 $(\bar{x},\bar{y})$。简要说明为何由对称性可知 $\bar{x}=\bar{y}$，并验证数值结果与此一致。[2]