Companion to the University-Style Practice Set大学风格练习题配套解析
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
(a) Evaluate $\iint_{R}(3x^{2}y+2y)\,dA$ over $R=[0,2]\times[1,3]$, integrating $x$ first; (b) for $g(x,y)=xe^{xy}$, write both iterated integrals and evaluate the tractable one.(a) 先对 $x$ 积分，计算 $R=[0,2]\times[1,3]$ 上的 $\iint_{R}(3x^{2}y+2y)\,dA$；(b) 对 $g(x,y)=xe^{xy}$，写出两个累次积分，并计算可化简的那个。
Fubini's theorem states that for a continuous function on a rectangle, the double integral equals either iterated integral. Integrating $x$ first over $[0,2]$: (M1)Fubini 定理指出，对于矩形上的连续函数，二重积分等于任意一种累次积分。先对 $[0,2]$ 上的 $x$ 积分：(M1)
$$\iint_{R}(3x^{2}y+2y)\,dA = \int_{1}^{3}\!\int_{0}^{2}(3x^{2}y+2y)\,dx\,dy = \int_{1}^{3}\Bigl[x^{3}y+2xy\Bigr]_{0}^{2}dy.$$Evaluating the inner integral: $[x^{3}y+2xy]_0^2 = 8y+4y=12y$. (A1) Now integrate over $y$: (M1)计算内层积分：$[x^{3}y+2xy]_0^2 = 8y+4y=12y$。(A1) 再对 $y$ 积分：(M1)
$$\int_{1}^{3}12y\,dy = \Bigl[6y^{2}\Bigr]_{1}^{3} = 6(9)-6(1) = 54-6 = 48.$$(A1)
Fubini guarantees (M1)Fubini 定理保证 (M1)
$$\int_{0}^{2}\!\int_{1}^{3}xe^{xy}\,dy\,dx = \int_{1}^{3}\!\int_{0}^{2}xe^{xy}\,dx\,dy.$$The $dy\,dx$ order has inner integral $\int_{1}^{3}xe^{xy}\,dy = \bigl[e^{xy}\bigr]_{y=1}^{y=3}=e^{3x}-e^{x}$, which is elementary. (A1) Integrating over $x$: (M1)$dy\,dx$ 次序的内层积分为 $\int_{1}^{3}xe^{xy}\,dy = \bigl[e^{xy}\bigr]_{y=1}^{y=3}=e^{3x}-e^{x}$，可化为初等形式。(A1) 再对 $x$ 积分：(M1)
$$\int_{0}^{2}(e^{3x}-e^{x})\,dx = \Bigl[\tfrac{1}{3}e^{3x}-e^{x}\Bigr]_{0}^{2} = \tfrac{1}{3}e^{6}-e^{2} - \bigl(\tfrac{1}{3}-1\bigr) = \tfrac{1}{3}(e^{6}-1)-(e^{2}-1).$$(A1) (The $dx\,dy$ order requires integrating $xe^{xy}$ with respect to $x$, which does not simplify in closed form over a fixed rectangle.)(A1)（$dx\,dy$ 次序需对 $x$ 积分 $xe^{xy}$，在固定矩形上无法化为初等闭合形式。）
$D$ bounded by $y=x^{2}$ and $y=2x$: (a) Type I setup for $\iint_{D}(x+y)\,dA$; (b) evaluate; (c) Type II form (do not evaluate).$D$ 由 $y=x^{2}$ 与 $y=2x$ 围成：(a) 将 $\iint_{D}(x+y)\,dA$ 设置为第 I 型；(b) 计算；(c) 写出第 II 型形式（无需计算）。
Find intersections: $x^{2}=2x$ gives $x(x-2)=0$, so $x=0$ and $x=2$. (M1) For $0\le x\le2$, the parabola $y=x^{2}$ lies below the line $y=2x$, so $D$ is the Type I region $0\le x\le2$, $x^{2}\le y\le2x$. (A1) The iterated integral is求交点：$x^{2}=2x$ 给出 $x(x-2)=0$，即 $x=0$ 与 $x=2$。(M1) 当 $0\le x\le2$ 时，抛物线 $y=x^{2}$ 在直线 $y=2x$ 下方，故 $D$ 为第 I 型区域，$0\le x\le2$，$x^{2}\le y\le2x$。(A1) 累次积分为
$$\iint_{D}(x+y)\,dA = \int_{0}^{2}\!\int_{x^{2}}^{2x}(x+y)\,dy\,dx. \quad \text{(A1)}$$Inner integral over $y$: (M1)对 $y$ 的内层积分：(M1)
$$\int_{x^{2}}^{2x}(x+y)\,dy = \Bigl[xy+\tfrac{y^{2}}{2}\Bigr]_{x^{2}}^{2x} = \bigl(2x^{2}+2x^{2}\bigr)-\bigl(x^{3}+\tfrac{x^{4}}{2}\bigr) = 4x^{2}-x^{3}-\tfrac{x^{4}}{2}.$$(A1) Outer integral: (M1)(A1) 外层积分：(M1)
$$\int_{0}^{2}\!\Bigl(4x^{2}-x^{3}-\tfrac{x^{4}}{2}\Bigr)\,dx = \Bigl[\tfrac{4x^{3}}{3}-\tfrac{x^{4}}{4}-\tfrac{x^{5}}{10}\Bigr]_{0}^{2} = \tfrac{32}{3}-4-\tfrac{32}{10}.$$Converting to a common denominator of 30: $\tfrac{320}{30}-\tfrac{120}{30}-\tfrac{96}{30}=\tfrac{104}{30}=\dfrac{52}{15}$. (A1)通分为 30：$\tfrac{320}{30}-\tfrac{120}{30}-\tfrac{96}{30}=\tfrac{104}{30}=\dfrac{52}{15}$。(A1)
The $y$-range is $0\le y\le4$ (the line meets the parabola at $(0,0)$ and $(2,4)$). (M1) For fixed $y$, $x$ runs from the line $x=y/2$ on the left to the parabola $x=\sqrt{y}$ on the right:$y$ 的范围为 $0\le y\le4$（直线与抛物线交于 $(0,0)$ 和 $(2,4)$）。(M1) 对固定的 $y$，$x$ 从左侧直线 $x=y/2$ 延伸至右侧抛物线 $x=\sqrt{y}$：
$$\iint_{D}(x+y)\,dA = \int_{0}^{4}\!\int_{y/2}^{\sqrt{y}}(x+y)\,dx\,dy. \quad \text{(A1)}$$Reverse the order of $\int_{0}^{1}\!\int_{x}^{1}e^{y^{2}}\,dy\,dx$ and evaluate.交换 $\int_{0}^{1}\!\int_{x}^{1}e^{y^{2}}\,dy\,dx$ 的积分次序并计算。
As given, the outer integral is $x\in[0,1]$ and for each $x$ the inner integral runs $y\in[x,1]$. (M1) This is the triangular region above the diagonal $y=x$ in the unit square, bounded by $x=0$, $y=1$, and $y=x$. To swap: fix $y\in[0,1]$; for each $y$, $x$ ranges from $0$ to $y$ (the left side of the triangle). (A1) The reversed integral is如所给，外层积分为 $x\in[0,1]$，对每个 $x$ 内层积分为 $y\in[x,1]$。(M1) 这是单位正方形内对角线 $y=x$ 上方的三角形区域，由 $x=0$、$y=1$ 和 $y=x$ 围成。交换次序：固定 $y\in[0,1]$；对每个 $y$，$x$ 从 $0$ 到 $y$（三角形左侧）。(A1) 交换后的积分为
$$\int_{0}^{1}\!\int_{x}^{1}e^{y^{2}}\,dy\,dx = \int_{0}^{1}\!\int_{0}^{y}e^{y^{2}}\,dx\,dy. \quad \text{(A1)}$$In the reversed integral, $e^{y^{2}}$ is constant with respect to $x$: (M1)在交换后的积分中，$e^{y^{2}}$ 关于 $x$ 为常数：(M1)
$$\int_{0}^{1}\!\int_{0}^{y}e^{y^{2}}\,dx\,dy = \int_{0}^{1}e^{y^{2}}\cdot y\,dy.$$(A1 for the inner integral giving the factor $y$) Substituting $u=y^{2}$, $du=2y\,dy$: (M1)（内层积分给出因子 $y$，得 A1）令 $u=y^{2}$，$du=2y\,dy$：(M1)
$$\int_{0}^{1}ye^{y^{2}}\,dy = \tfrac{1}{2}\int_{0}^{1}e^{u}\,du = \tfrac{1}{2}\bigl[e^{u}\bigr]_{0}^{1} = \tfrac{1}{2}(e-1) = \dfrac{e-1}{2}. \quad \text{(A1)}$$Evaluate $\iint_{D}(x^{2}+y^{2})^{3/2}\,dA$ over the disk $D: x^{2}+y^{2}\le9$.计算圆盘 $D: x^{2}+y^{2}\le9$ 上的 $\iint_{D}(x^{2}+y^{2})^{3/2}\,dA$。
Set $x=r\cos\theta$, $y=r\sin\theta$, $dA=r\,dr\,d\theta$. The disk $x^{2}+y^{2}\le9$ becomes $0\le r\le3$, $0\le\theta\le2\pi$. (M1) The integrand simplifies: $(x^{2}+y^{2})^{3/2}=(r^{2})^{3/2}=r^{3}$. (A1) Including the area element:令 $x=r\cos\theta$，$y=r\sin\theta$，$dA=r\,dr\,d\theta$。圆盘 $x^{2}+y^{2}\le9$ 变为 $0\le r\le3$，$0\le\theta\le2\pi$。(M1) 被积函数化简为 $(x^{2}+y^{2})^{3/2}=(r^{2})^{3/2}=r^{3}$。(A1) 代入面积元：
$$\iint_{D}(x^{2}+y^{2})^{3/2}\,dA = \int_{0}^{2\pi}\!\int_{0}^{3}r^{3}\cdot r\,dr\,d\theta = \int_{0}^{2\pi}\!\int_{0}^{3}r^{4}\,dr\,d\theta. \quad \text{(A1)}$$The integrand separates in $r$ and $\theta$: (M1)被积函数关于 $r$ 和 $\theta$ 可分离：(M1)
$$\int_{0}^{2\pi}d\theta\cdot\int_{0}^{3}r^{4}\,dr = 2\pi\cdot\Bigl[\tfrac{r^{5}}{5}\Bigr]_{0}^{3} = 2\pi\cdot\dfrac{243}{5} = \dfrac{486\pi}{5}. \quad \text{(A1·A1)}$$For $x=r\cos\theta$, $y=r\sin\theta$: (a) compute $\partial(x,y)/\partial(r,\theta)$; (b) justify $dA=r\,dr\,d\theta$; (c) give the correct area of $r\le a$ and describe the error from omitting $r$.对 $x=r\cos\theta$，$y=r\sin\theta$：(a) 计算 $\partial(x,y)/\partial(r,\theta)$；(b) 说明 $dA=r\,dr\,d\theta$ 的依据；(c) 给出 $r\le a$ 的正确面积，并描述省略 $r$ 导致的错误。
The Jacobian of the transformation $(r,\theta)\mapsto(x,y)$ is the determinant (M1)变换 $(r,\theta)\mapsto(x,y)$ 的 Jacobian 为行列式 (M1)
$$\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix}\dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial\theta}\\[8pt]\dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial\theta}\end{vmatrix} = \begin{vmatrix}\cos\theta & -r\sin\theta\\\sin\theta & r\cos\theta\end{vmatrix}.$$Computing: $\cos\theta\cdot r\cos\theta - (-r\sin\theta)\cdot\sin\theta = r\cos^{2}\theta + r\sin^{2}\theta$. (M1·A1) Since $\cos^{2}\theta+\sin^{2}\theta=1$:计算：$\cos\theta\cdot r\cos\theta - (-r\sin\theta)\cdot\sin\theta = r\cos^{2}\theta + r\sin^{2}\theta$。(M1·A1) 由于 $\cos^{2}\theta+\sin^{2}\theta=1$：
$$\frac{\partial(x,y)}{\partial(r,\theta)} = r. \quad \text{(A1)}$$The general change-of-variables theorem states $dA=\bigl|\partial(x,y)/\partial(r,\theta)\bigr|\,dr\,d\theta$. (R1) Since $r\ge0$ we have $|r|=r$, so $dA=r\,dr\,d\theta$. The factor $r$ is non-negative for all admissible $r$, so no absolute value sign is needed in practice. (A1)一般换元定理指出 $dA=\bigl|\partial(x,y)/\partial(r,\theta)\bigr|\,dr\,d\theta$。(R1) 由于 $r\ge0$，有 $|r|=r$，故 $dA=r\,dr\,d\theta$。对所有允许的 $r$，因子 $r$ 均非负，实践中无需绝对值符号。(A1)
With the correct area element: (M1)使用正确的面积元：(M1)
$$\text{Area} = \int_{0}^{2\pi}\!\int_{0}^{a}r\,dr\,d\theta = 2\pi\cdot\Bigl[\tfrac{r^{2}}{2}\Bigr]_{0}^{a} = 2\pi\cdot\tfrac{a^{2}}{2} = \pi a^{2}. \quad \text{(A1)}$$If the student writes $dA=dr\,d\theta$ (omitting the $r$), they obtain $\int_0^{2\pi}\int_0^a dr\,d\theta = 2\pi a$, which is a length, not an area. The answer is dimensionally wrong and equals neither $\pi a^2$ nor anything geometrically meaningful.若学生写出 $dA=dr\,d\theta$（省略了 $r$），则得到 $\int_0^{2\pi}\int_0^a dr\,d\theta = 2\pi a$，这是一个长度而非面积。答案量纲有误，既不等于 $\pi a^2$，也没有任何几何意义。
$E$: tetrahedron with $x\ge0$, $y\ge0$, $z\ge0$, $x+y+z\le1$. (a) Set up $\iiint_E dV$ in order $dz\,dy\,dx$; (b) evaluate and confirm $V=\frac{1}{6}$.$E$：满足 $x\ge0$，$y\ge0$，$z\ge0$，$x+y+z\le1$ 的四面体。(a) 按 $dz\,dy\,dx$ 次序建立 $\iiint_E dV$；(b) 计算并验证 $V=\frac{1}{6}$。
Describe the region from the inside out. For fixed $(x,y)$, the constraint $z\le1-x-y$ with $z\ge0$ gives $0\le z\le1-x-y$, which requires $1-x-y\ge0$, i.e. $y\le1-x$. (M1) For fixed $x$, $y$ ranges over $0\le y\le1-x$. (A1) Finally $x$ ranges over $0\le x\le1$: (M1)由内向外描述区域。对固定的 $(x,y)$，约束 $z\le1-x-y$ 与 $z\ge0$ 给出 $0\le z\le1-x-y$，这要求 $1-x-y\ge0$，即 $y\le1-x$。(M1) 对固定的 $x$，$y$ 的范围为 $0\le y\le1-x$。(A1) 最终 $x$ 的范围为 $0\le x\le1$：(M1)
$$\iiint_{E}dV = \int_{0}^{1}\!\int_{0}^{1-x}\!\int_{0}^{1-x-y}dz\,dy\,dx. \quad \text{(A1)}$$Innermost integral: $\int_{0}^{1-x-y}dz = 1-x-y$. (M1) Middle integral: (A1)最内层积分：$\int_{0}^{1-x-y}dz = 1-x-y$。(M1) 中间积分：(A1)
$$\int_{0}^{1-x}(1-x-y)\,dy = \Bigl[(1-x)y-\tfrac{y^{2}}{2}\Bigr]_{0}^{1-x} = (1-x)^{2}-\tfrac{(1-x)^{2}}{2} = \tfrac{(1-x)^{2}}{2}.$$(M1) Outer integral:(M1) 外层积分：
$$\int_{0}^{1}\tfrac{(1-x)^{2}}{2}\,dx = \tfrac{1}{2}\Bigl[-\tfrac{(1-x)^{3}}{3}\Bigr]_{0}^{1} = \tfrac{1}{2}\cdot\tfrac{1}{3} = \dfrac{1}{6}. \quad \text{(A1)}$$Substitution $u=\tfrac{1}{5}(2x-y)$, $v=\tfrac{1}{5}(3y-x)$ maps parallelogram $D$ to $S=[0,1]^{2}$. (a) Invert to find $x(u,v)$, $y(u,v)$; (b) compute the Jacobian; (c) evaluate $\iint_D(2x-y)\,dA$.换元 $u=\tfrac{1}{5}(2x-y)$，$v=\tfrac{1}{5}(3y-x)$ 将平行四边形 $D$ 映射至 $S=[0,1]^{2}$。(a) 求逆，给出 $x(u,v)$，$y(u,v)$；(b) 计算 Jacobian；(c) 计算 $\iint_D(2x-y)\,dA$。
From the given substitution: $2x-y=5u$ and $-x+3y=5v$. (M1) Multiply the first equation by 3 and add the second: $6x-3y-x+3y=15u+5v$, giving $5x=15u+5v$, so $x=3u+v$. (A1) Substitute back: $y=2x-5u=2(3u+v)-5u=u+2v$. (A1)由所给换元：$2x-y=5u$ 且 $-x+3y=5v$。(M1) 将第一个方程乘以 3 后加第二个方程：$6x-3y-x+3y=15u+5v$，得 $5x=15u+5v$，即 $x=3u+v$。(A1) 回代：$y=2x-5u=2(3u+v)-5u=u+2v$。(A1)
With $x=3u+v$ and $y=u+2v$: (M1)由 $x=3u+v$ 和 $y=u+2v$：(M1)
$$\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\[8pt]\dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}\end{vmatrix} = \begin{vmatrix}3 & 1\\1 & 2\end{vmatrix} = 3\cdot2 - 1\cdot1 = 5. \quad \text{(A1·A1)}$$The integrand $2x-y=5u$ in the new variables. (M1) The change-of-variables formula gives:新变量下被积函数 $2x-y=5u$。(M1) 变量替换公式给出：
$$\iint_{D}(2x-y)\,dA = \iint_{S}5u\cdot\bigl|\tfrac{\partial(x,y)}{\partial(u,v)}\bigr|\,du\,dv = \iint_{S}5u\cdot5\,du\,dv = 25\int_{0}^{1}\!\int_{0}^{1}u\,du\,dv. \quad \text{(A1)}$$Evaluating: (M1)计算：(M1)
$$25\int_{0}^{1}dv\cdot\int_{0}^{1}u\,du = 25\cdot1\cdot\tfrac{1}{2} = \dfrac{25}{2}. \quad \text{(A1)}$$$E$: above $z=\sqrt{x^{2}+y^{2}}$, inside $x^{2}+y^{2}+z^{2}=4$. Find $V$ using spherical coordinates.$E$：位于 $z=\sqrt{x^{2}+y^{2}}$ 上方、$x^{2}+y^{2}+z^{2}=4$ 内部。用球坐标求体积 $V$。
In spherical coordinates: $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, $z=\rho\cos\phi$, and $x^{2}+y^{2}+z^{2}=\rho^{2}$. (M1)在球坐标中：$x=\rho\sin\phi\cos\theta$，$y=\rho\sin\phi\sin\theta$，$z=\rho\cos\phi$，且 $x^{2}+y^{2}+z^{2}=\rho^{2}$。(M1)
The sphere $x^{2}+y^{2}+z^{2}=4$ becomes $\rho=2$. The cone $z=\sqrt{x^{2}+y^{2}}$: the left side is $\rho\cos\phi$ and the right side is $\sqrt{\rho^{2}\sin^{2}\phi}=\rho\sin\phi$ (since $\rho\ge0$). So the cone is $\cos\phi=\sin\phi$, i.e. $\tan\phi=1$, giving $\phi=\pi/4$. (A1) The solid $E$ lies above the cone and inside the sphere, i.e. where $\phi$ is small (near the north pole). Thus: (M1)球面 $x^{2}+y^{2}+z^{2}=4$ 变为 $\rho=2$。锥面 $z=\sqrt{x^{2}+y^{2}}$：左侧为 $\rho\cos\phi$，右侧为 $\sqrt{\rho^{2}\sin^{2}\phi}=\rho\sin\phi$（因为 $\rho\ge0$）。故锥面方程为 $\cos\phi=\sin\phi$，即 $\tan\phi=1$，给出 $\phi=\pi/4$。(A1) 立体 $E$ 位于锥面上方、球面内部，即 $\phi$ 较小（靠近北极）的区域。因此：(M1)
$$0\le\theta\le2\pi,\quad 0\le\phi\le\frac{\pi}{4},\quad 0\le\rho\le2. \quad \text{(A1)}$$The volume element in spherical coordinates is $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$. (A1)球坐标中的体积元为 $dV=\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta$。(A1)
$$\iiint_{E}dV = \int_{0}^{2\pi}\!\int_{0}^{\pi/4}\!\int_{0}^{2}\rho^{2}\sin\phi\,d\rho\,d\phi\,d\theta. \quad \text{(A1)}$$The integral separates into three factors: (M1)积分分离为三个因子：(M1)
$$\int_{0}^{2\pi}d\theta\cdot\int_{0}^{\pi/4}\sin\phi\,d\phi\cdot\int_{0}^{2}\rho^{2}\,d\rho.$$Evaluate each factor: $\int_0^{2\pi}d\theta=2\pi$; $\int_0^{\pi/4}\sin\phi\,d\phi=[-\cos\phi]_0^{\pi/4}=1-\dfrac{\sqrt{2}}{2}=\dfrac{2-\sqrt{2}}{2}$; $\int_0^{2}\rho^{2}\,d\rho=\dfrac{8}{3}$. (A1) Combining: (M1)计算各因子：$\int_0^{2\pi}d\theta=2\pi$；$\int_0^{\pi/4}\sin\phi\,d\phi=[-\cos\phi]_0^{\pi/4}=1-\dfrac{\sqrt{2}}{2}=\dfrac{2-\sqrt{2}}{2}$；$\int_0^{2}\rho^{2}\,d\rho=\dfrac{8}{3}$。(A1) 合并：(M1)
$$V = 2\pi\cdot\frac{2-\sqrt{2}}{2}\cdot\frac{8}{3} = \frac{8\pi(2-\sqrt{2})}{3}. \quad \text{(A1)}$$$E$: between $z=0$ and $z=9-x^{2}-y^{2}$ (with $z\ge0$). Find $V$ using cylindrical coordinates.$E$：夹在 $z=0$ 与 $z=9-x^{2}-y^{2}$ 之间（$z\ge0$）。用柱坐标求 $V$。
In cylindrical coordinates: $x=r\cos\theta$, $y=r\sin\theta$, $z=z$, and $dV=r\,dz\,dr\,d\theta$. The paraboloid becomes $z=9-r^{2}$; the region $z\ge0$ requires $9-r^{2}\ge0$, i.e. $r\le3$. (M1) The solid has full circular symmetry, so $\theta$ ranges over $[0,2\pi]$. For each $(r,\theta)$, $z$ runs from $0$ to $9-r^{2}$: (A1)在柱坐标中：$x=r\cos\theta$，$y=r\sin\theta$，$z=z$，体积元 $dV=r\,dz\,dr\,d\theta$。抛物面变为 $z=9-r^{2}$；条件 $z\ge0$ 要求 $9-r^{2}\ge0$，即 $r\le3$。(M1) 立体具有完全的圆形对称性，故 $\theta$ 遍历 $[0,2\pi]$。对每个 $(r,\theta)$，$z$ 从 $0$ 到 $9-r^{2}$：(A1)
$$0\le\theta\le2\pi,\quad 0\le r\le3,\quad 0\le z\le9-r^{2}. \quad \text{(A1)}$$Write the integral: (M1)写出积分：(M1)
$$V = \int_{0}^{2\pi}\!\int_{0}^{3}\!\int_{0}^{9-r^{2}}r\,dz\,dr\,d\theta.$$Innermost integral: $\int_0^{9-r^2}dz=9-r^{2}$. (A1) The $\theta$ integral contributes $2\pi$. Middle integral: (M1)最内层积分：$\int_0^{9-r^2}dz=9-r^{2}$。(A1) $\theta$ 积分贡献 $2\pi$。中间积分：(M1)
$$2\pi\int_{0}^{3}r(9-r^{2})\,dr = 2\pi\int_{0}^{3}(9r-r^{3})\,dr = 2\pi\Bigl[\tfrac{9r^{2}}{2}-\tfrac{r^{4}}{4}\Bigr]_{0}^{3}.$$Evaluating: $\tfrac{9\cdot9}{2}-\tfrac{81}{4}=\tfrac{81}{2}-\tfrac{81}{4}=\tfrac{81}{4}$. (A1) Therefore $V=2\pi\cdot\dfrac{81}{4}=\dfrac{81\pi}{2}$. (A1)计算：$\tfrac{9\cdot9}{2}-\tfrac{81}{4}=\tfrac{81}{2}-\tfrac{81}{4}=\tfrac{81}{4}$。(A1) 因此 $V=2\pi\cdot\dfrac{81}{4}=\dfrac{81\pi}{2}$。(A1)
Quarter-disk $D$ in the first quadrant, $x^{2}+y^{2}\le1$, density $\delta(x,y)=x+y$. (a) Find mass $m$; (b) find $M_y$ and $M_x$; (c) find centroid $(\bar x,\bar y)$ and confirm $\bar x=\bar y$ by symmetry.第一象限四分之一圆盘 $D$，$x^{2}+y^{2}\le1$，密度 $\delta(x,y)=x+y$。(a) 求质量 $m$；(b) 求 $M_y$ 和 $M_x$；(c) 求质心 $(\bar x,\bar y)$ 并由对称性验证 $\bar x=\bar y$。
In polar: $x+y=r(\cos\theta+\sin\theta)$, $dA=r\,dr\,d\theta$. The quarter-disk is $0\le r\le1$, $0\le\theta\le\pi/2$. (M1)在极坐标中：$x+y=r(\cos\theta+\sin\theta)$，$dA=r\,dr\,d\theta$。四分之一圆盘为 $0\le r\le1$，$0\le\theta\le\pi/2$。(M1)
$$m = \int_{0}^{\pi/2}\!\int_{0}^{1}r(\cos\theta+\sin\theta)\cdot r\,dr\,d\theta = \int_{0}^{\pi/2}(\cos\theta+\sin\theta)\,d\theta\cdot\int_{0}^{1}r^{2}\,dr.$$(A1 for the separation) $\int_0^{\pi/2}(\cos\theta+\sin\theta)\,d\theta=[\sin\theta-\cos\theta]_0^{\pi/2}=(1-0)-(0-1)=2$. (M1) $\int_0^{1}r^{2}\,dr=\tfrac{1}{3}$. Therefore $m=2\cdot\tfrac{1}{3}=\dfrac{2}{3}$. (A1)（变量分离得 A1）$\int_0^{\pi/2}(\cos\theta+\sin\theta)\,d\theta=[\sin\theta-\cos\theta]_0^{\pi/2}=(1-0)-(0-1)=2$。(M1) $\int_0^{1}r^{2}\,dr=\tfrac{1}{3}$。因此 $m=2\cdot\tfrac{1}{3}=\dfrac{2}{3}$。(A1)
For $M_y=\iint_D x\,\delta\,dA=\iint_D x(x+y)\,dA$. In polar, $x=r\cos\theta$ and $x(x+y)=r^{2}\cos\theta(\cos\theta+\sin\theta)$: (M1)对 $M_y=\iint_D x\,\delta\,dA=\iint_D x(x+y)\,dA$，在极坐标中 $x=r\cos\theta$，$x(x+y)=r^{2}\cos\theta(\cos\theta+\sin\theta)$：(M1)
$$M_{y} = \int_{0}^{\pi/2}\cos\theta(\cos\theta+\sin\theta)\,d\theta\cdot\int_{0}^{1}r^{3}\,dr.$$The $r$-integral: $\int_0^1 r^3\,dr=\tfrac{1}{4}$. The $\theta$-integral: $\int_0^{\pi/2}(\cos^{2}\theta+\sin\theta\cos\theta)\,d\theta = \tfrac{\pi}{4}+\tfrac{1}{2}=\tfrac{\pi+2}{4}$. (A1) So $M_{y}=\tfrac{1}{4}\cdot\tfrac{\pi+2}{4}=\dfrac{\pi+2}{16}$.$r$ 积分：$\int_0^1 r^3\,dr=\tfrac{1}{4}$。$\theta$ 积分：$\int_0^{\pi/2}(\cos^{2}\theta+\sin\theta\cos\theta)\,d\theta = \tfrac{\pi}{4}+\tfrac{1}{2}=\tfrac{\pi+2}{4}$。(A1) 故 $M_{y}=\tfrac{1}{4}\cdot\tfrac{\pi+2}{4}=\dfrac{\pi+2}{16}$。
By the symmetry argument in (c) below, $M_x=M_y=\dfrac{\pi+2}{16}$ (verified by analogous computation with $\sin\theta$ and $\cos\theta$ swapped). (M1·A1)由下方 (c) 的对称性论证，$M_x=M_y=\dfrac{\pi+2}{16}$（通过将 $\sin\theta$ 与 $\cos\theta$ 互换做类似计算验证）。(M1·A1)
The centroid is $\bar{x}=M_y/m$ and $\bar{y}=M_x/m$: (A1)质心为 $\bar{x}=M_y/m$，$\bar{y}=M_x/m$：(A1)
$$\bar{x} = \bar{y} = \frac{(\pi+2)/16}{2/3} = \frac{\pi+2}{16}\cdot\frac{3}{2} = \frac{3(\pi+2)}{32}.$$Symmetry argument: the region $D$ and the density $\delta(x,y)=x+y$ are both symmetric under swapping $x$ and $y$ (i.e. reflecting across the line $y=x$). This symmetry exchanges $M_y$ and $M_x$ while preserving $m$, so $M_y=M_x$ and hence $\bar{x}=\bar{y}$. (A1)对称性论证：区域 $D$ 和密度 $\delta(x,y)=x+y$ 在交换 $x$ 与 $y$（即关于直线 $y=x$ 作反射）时均保持对称。此对称性将 $M_y$ 与 $M_x$ 互换而不改变 $m$，故 $M_y=M_x$，即 $\bar{x}=\bar{y}$。(A1)