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微积分3:多重积分

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例子问题

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问题1:柱坐标下的三重积分

评估 $\int \int \int_D y \ dV$ ,在那里是平面下面的区域吗 z=x+1 ,高于平面和柱体之间 x^2+y^2=1 , x^2+y^2=9 ．

可能的答案:

$2 \pi$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\pi$

正确答案:

解释：

我们需要算出积分的边界。

我们需要把所有东西都转换成柱坐标。记住，我们是在Plane，意思是我们在上面 z=0 ．

$0\leq z \leq x+1 \rightarrow 0\leq z \leq r\cos(\theta)+1$

该地区在两个圆之间 x^2+y^2=1 , x^2+y^2=9 ．

这意味着

$0 \leq \theta \leq 2\pi$

$1\leq r \leq 3$

$\int \int \int_D y \ dV=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r\sin(\theta) r\ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} \int_{0}^{r\cos(\theta)+1} r^2\sin(\theta) \ dz \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} zr^2\sin(\theta) \Big|_{0}^{r\cos(\theta)+1} \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} (r\cos(\theta)+1)r^2\sin(\theta)-0 \ dr \ d\theta$

$=\int_{0}^{2\pi} \int_{1}^{3} r^3\sin(\theta)\cos(\theta)+r^2\sin(\theta) \ dr \ d\theta$

$=\int_{0}^{2\pi} \frac{1}{4}r^4\sin(\theta)\cos(\theta)+\frac{1}{3}r^3\sin(\theta) \Big|_{1}^{3}\ d\theta$

$=\int_{0}^{2\pi} (\frac{1}{4}(3)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(3)^3\sin(\theta))-(\frac{1}{4}(1)^4\sin(\theta)\cos(\theta)+\frac{1}{3}(1)^3\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} (\frac{81}{4}\sin(\theta)\cos(\theta)+9\sin(\theta))-(\frac{1}{4}\sin(\theta)\cos(\theta)+\frac{1}{3}\sin(\theta)) \ d\theta$

$=\int_{0}^{2\pi} 20\sin(\theta)\cos(\theta)+\frac{26}{3}\sin(\theta) \ d\theta$

$= -\frac{1}{2}\cdot20\cos^2(\theta)-\frac{26}{3}\cos(\theta)\Big|_{0}^{2\pi}$

$= -\frac{1}{2}\cdot20\cos^2(2\pi)-\frac{26}{3}\cos(2\pi)-(-\frac{1}{2}\cdot20\cos^2(0)-\frac{26}{3}\cos(0))$

$= -\frac{1}{2}\cdot20(1)-\frac{26}{3}(1)-(-\frac{1}{2}\cdot20(1)-\frac{26}{3}(1))$

$= -10-\frac{26}{3}-(-10-\frac{26}{3})=0$

报告一个错误

问题2:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.23\text{ and }1.56\\&\text{and length }1.97\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-0.07

-0.02

0.03

0.01

正确答案:

0.03

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (3cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ (sin(z + 1)cos(x^{2} + y^{2}))}{25})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{-0.094})=0.53\pi;\theta_2=arctan(\frac{-0.771}{0.637})=1.72\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\end{align*}$

$\begin{align*}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}\int_{0.23}^{1.56}(-\frac{ (3\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (rcos(r^{2})sin(z + 1))}{25})drd\theta dz=(-\frac{ sin(\frac{(3\cdot r^{2})}{2})}{41}-\frac{ (sin(r^{2})sin(z + 1))}{50})d\theta dz|_{0.23}^{1.56}\\&\int_{0}^{1.97}\int_{0.53\pi}^{1.72\pi}(0.01381 - 0.01195sin(z + 1))d\theta dz=(-1.0\theta\cdot (0.01195sin(z + 1) - 0.01381))dz|_{0.53\pi}^{1.72\pi}\\&\int_{0}^{1.97}(0.05165 - 0.04467sin(z + 1))dz=0.03\end{align*}$

报告一个错误

问题3:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.92\\&\text{and length }1.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

$-2.08871\cdot10^{15}$

$8.35483\cdot10^{15}$

$-8.35483\cdot10^{15}$

$1.39247\cdot10^{15}$

正确答案:

$-8.35483\cdot10^{15}$

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}-\frac{ (49e^{(20z)}e^{(x^{2} + y^{2})})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}\int_{0}^{1.92}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ (49\cdot re^{(r^{2})}e^{(20z)})}{5})drd\theta dz=(-\frac{ (49e^{(20z + r^{2})})}{10}-\frac{ (3e^{(-\frac{(2\cdot r^{2})}{3})})}{16})d\theta dz|_{0}^{1.92}\\&\int_{0}^{1.65}\int_{0.43\pi}^{1.73\pi}(0.1714 - 190.6e^{(20z)})d\theta dz=(-1.0\theta\cdot (190.6e^{(20z)} - 0.1714))dz|_{0.43\pi}^{1.73\pi}\\&\int_{0}^{1.65}(0.7002 - 778.5e^{(20z)})dz=(0.7002z - 38.92e^{(20z)})|_{0}^{1.65}=-8.35483\cdot10^{15}\end{align*}$

报告一个错误

问题4:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(5cos(x^{2} + y^{2}) +\frac{ (3\cdot 2^ze^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{13})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.31\text{ and }1.89\\&\text{and length }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.988,0.156)\text{ and }\overrightarrow{u_2}=(-1.000,0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-27.28

-4.55

27.28

2.27

正确答案:

-4.55

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(5cos(x^{2} + y^{2}) +\frac{ (3\cdot 2^ze^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{13})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.156}{0.988})=0.05\pi;\theta_2=arctan(\frac{0.031}{-1.000})=0.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}\int_{0.31}^{1.89}(5\cdot rcos(r^{2}) +\frac{ (3\cdot 2^z\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{13})drd\theta dz=(\frac{(5sin(r^{2}))}{2}-\frac{ (2^ze^{(-\frac{(3\cdot r^{2})}{2})})}{13})d\theta dz|_{0.31}^{1.89}\\&\int_{0}^{1.31}\int_{0.05\pi}^{0.99\pi}(0.06623\cdot 2^z - 1.283)d\theta dz=(\theta\cdot (0.06623\cdot 2^z - 1.283))dz|_{0.05\pi}^{0.99\pi}\\&\int_{0}^{1.31}(0.1956\cdot 2^z - 3.789)dz=(0.2822\cdot 2^z - 3.789z)|_{0}^{1.31}=-4.55\end{align*}$

报告一个错误

问题5:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 16e^{(x^{2} + y^{2})} -\frac{ (5e^{(2z)})}{(19\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.42\text{ and }1.42\\&\text{and length }1.72\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.368,0.930)\text{ and }\overrightarrow{u_2}=(-0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-239.31

39.89

-1196.57

239.31

正确答案:

-239.31

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- 16e^{(x^{2} + y^{2})} -\frac{ (5e^{(2z)})}{(19\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.930}{-0.368})=0.62\pi;\theta_2=arctan(\frac{-0.988}{-0.156})=1.45\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}\int_{0.42}^{1.42}(- 16\cdot re^{(r^{2})} -\frac{ (5e^{(2z)})}{(19\cdot r)})drd\theta dz=(- 8e^{(r^{2})} -\frac{ (5e^{(2z)}ln(r))}{19})d\theta dz|_{0.42}^{1.42}\\&\int_{0}^{1.72}\int_{0.62\pi}^{1.45\pi}(- 0.3206e^{(2z)} - 50.55)d\theta dz=(-1.0\theta\cdot (0.3206e^{(2z)} + 50.55))dz|_{0.62\pi}^{1.45\pi}\\&\int_{0}^{1.72}(- 0.8359e^{(2z)} - 131.8)dz=(- 131.8z - 0.4179e^{(2z)})|_{0}^{1.72}=-239.31\end{align*}$

报告一个错误

问题6:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2\cdot 4^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7}-\frac{ (7\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.27\text{ and }1.86\\&\text{and length }1.09\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

100.58

16.76

-83.82

-4.19

正确答案:

16.76

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2\cdot 4^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7}-\frac{ (7\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}\int_{0.27}^{1.86}(\frac{(2\cdot 4^{(\frac{r^{2}}{2})}\cdot r)}{7}-\frac{ (7\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz=(\frac{2^{(r^{2})}}{(7ln(2))}-\frac{ (7\cdot 3^zsin(r^{2}))}{4})d\theta dz|_{0.27}^{1.86}\\&\int_{0}^{1.09}\int_{0.16\pi}^{1.62\pi}(0.6746\cdot 3^z + 2.051)d\theta dz=(\theta\cdot (0.6746\cdot 3^z + 2.051))dz|_{0.16\pi}^{1.62\pi}\\&\int_{0}^{1.09}(3.094\cdot 3^z + 9.405)dz=(9.405z + 2.817\cdot 3^z)|_{0}^{1.09}=16.76\end{align*}$

报告一个错误

问题7:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(- x^{2} - y^{2})})}{11}+\frac{ (49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.87\\&\text{and length }1.45\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.982,-0.187)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

211.22

-7.04

-42.24

7.04

正确答案:

-42.24

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(- x^{2} - y^{2})})}{11}+\frac{ (49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.187}{0.982})=1.94\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}\int_{0}^{1.87}(\frac{(3\cdot re^{(-r^{2})})}{11}+\frac{ (49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz=(\frac{(49sin(r^{2})e^{(2z)})}{12}-\frac{ (3e^{(-r^{2})})}{22})d\theta dz|_{0}^{1.87}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}(0.1322 - 1.421e^{(2z)})d\theta dz=(-\theta\cdot (1.421e^{(2z)} - 0.1322))dz|_{0.82\pi}^{1.94\pi}\\&\int_{0}^{1.45}(0.4653 - 4.998e^{(2z)})dz=(0.4653z - 2.499e^{(2z)})|_{0}^{1.45}=-42.24\end{align*}$

报告一个错误

问题8:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(48cos(3z)sin(2x^{2} + 2y^{2}) - 20sin(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.17\text{ and }1.61\\&\text{and length }0.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.125,-0.992)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

77.73

-6.48

-19.43

3.89

正确答案:

-19.43

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(48cos(3z)sin(2x^{2} + 2y^{2}) - 20sin(x^{2} + y^{2}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.992}{0.125})=1.54\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}\int_{0.17}^{1.61}(48\cdot rcos(3z)sin(2\cdot r^{2}) - 20\cdot rsin(r^{2}))drd\theta dz=(10cos(r^{2}) - 12cos(3z)cos(2\cdot r^{2}))d\theta dz|_{0.17}^{1.61}\\&\int_{0}^{0.54}\int_{0.75\pi}^{1.54\pi}(6.526cos(3z) - 18.52)d\theta dz=(\theta\cdot (6.526cos(3z) - 18.52))dz|_{0.75\pi}^{1.54\pi}\\&\int_{0}^{0.54}(16.2cos(3z) - 45.97)dz=(10.8cos(1.5z)sin(1.5z) - 45.97z)|_{0}^{0.54}=-19.43\end{align*}$

报告一个错误

问题9:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(13cos(2x^{2} + 2y^{2}))}{6}-\frac{ (2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{15})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.94\\&\text{and length }1.01\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(-0.844,0.536)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

40.78

3.4

-20.39

-10.2

正确答案:

-10.2

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(13cos(2x^{2} + 2y^{2}))}{6}-\frac{ (2e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)})}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{0.536}{-0.844})=0.82\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}\int_{0}^{1.94}(\frac{(13\cdot rcos(2\cdot r^{2}))}{6}-\frac{ (2\cdot re^{(-2z)}e^{(\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz=(\frac{(13sin(2\cdot r^{2}))}{24}-\frac{ (2e^{(\frac{(3\cdot r^{2})}{2}- 2z)})}{45})d\theta dz|_{0}^{1.94}\\&\int_{0}^{1.01}\int_{0.16\pi}^{0.82\pi}(0.513 - 12.53e^{(-2z)})d\theta dz=(-\theta\cdot (12.53e^{(-2z)} - 0.513))dz|_{0.16\pi}^{0.82\pi}\\&\int_{0}^{1.01}(1.064 - 25.99e^{(-2z)})dz=(1.064z + 12.99e^{(-2z)})|_{0}^{1.01}=-10.2\end{align*}$

报告一个错误

问题10:柱坐标下的三重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{29}{(3\cdot (2x^{2} + 2y^{2}))}-\frac{ (4sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-z)})}{3})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.33\text{ and }1.47\\&\text{and length }1.88\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.094,0.996)\text{ and }\overrightarrow{u_2}=(-0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-11.55

138.64

-138.64

34.66

正确答案:

34.66

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{29}{(3\cdot (2x^{2} + 2y^{2}))}-\frac{ (4sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-z)})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}\int_{0.33}^{1.47}(\frac{29}{(6\cdot r)}-\frac{ (4\cdot re^{(-z)}sin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta dz=(\frac{(29ln(r))}{6}+\frac{ (4e^{(-z)}cos(\frac{(3\cdot r^{2})}{2}))}{9})d\theta dz|_{0.33}^{1.47}\\&\int_{0}^{1.88}\int_{0.47\pi}^{1.33\pi}(7.221 - 0.8808e^{(-z)})d\theta dz=(-\theta\cdot (0.8808e^{(-z)} - 7.221))dz|_{0.47\pi}^{1.33\pi}\\&\int_{0}^{1.88}(19.51 - 2.38e^{(-z)})dz=(19.51z + 2.38e^{(-z)})|_{0}^{1.88}=34.66\end{align*}$

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