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微积分3:二重积分

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6诊断测试 373年实践测试每日问题抽认卡学习的概念

例子问题

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例子问题1:二重积分

通过转换到极坐标来计算下面的积分。

$\int \int_D 10xy \ dA$ ,在那里圆与圆之间的半径是多少而且在第一象限。

可能的答案:

1000

1218

2436

2000

正确答案:

1218

解释：

我们要记住如何将笛卡尔坐标转化为极坐标。

$x=r\cos(\theta)$

$y=r\sin(\theta)$

让我们写出变量的范围而且 $\theta$ ．

$4\leq r \leq 10$

$0\leq \theta \leq \frac{\pi}{2}$

现在我们来建立二重积分，不要忘了额外的．

$\int_{0}^{\frac{\pi}{2}} \int_{4}^{10}(r\cos(\theta))(r\sin(\theta)) r\ dr\ d\theta$

$=\int_{0}^{\frac{\pi}{2}} \int_{4}^{10}r^3\cos(\theta)\sin(\theta)\ dr\ d\theta$

$=\int_{0}^{\frac{\pi}{2}} \frac{1}{4}r^4\cos(\theta)\sin(\theta)\Big|_{4}^{10}\ d\theta$

$=\int_{0}^{\frac{\pi}{2}} \frac{1}{4}(10)^4\cos(\theta)\sin(\theta)-\frac{1}{4}(4)^4\cos(\theta)\sin(\theta) d\theta$

$=\int_{0}^{\frac{\pi}{2}} 2500\cos(\theta)\sin(\theta)-64\cos(\theta)\sin(\theta) d\theta$

$=\int_{0}^{\frac{\pi}{2}} 2436\cos(\theta)\sin(\theta) d\theta$

$=-\frac{1}{2}\cdot 2436\cos^2(\theta) \Big|_{0}^{\frac{\pi}{2}}$

$=-\frac{1}{2}\cdot 2436\cos^2\Big(\Big(\frac{\pi}{2}\Big)\Big)+\frac{1}{2}\cdot 2436\cos^2\Big(\Big(0\Big)\Big)$

$=\frac{1}{2}\cdot 2436(1)$

=1218

报告一个错误

例子问题2:极坐标下的二重积分

评估积分

$\int\int_D(x^2+y^2)dxdy$

其中D是x轴上方以半径2为圆心的圆内的区域。

可能的答案:

$2\pi$

$4\pi$

$8\pi$

$\frac{8}{3}\pi$

$\frac{5}{4}\pi$

正确答案:

$4\pi$

解释：

笛卡尔坐标到极坐标的转换是:

$x=r\cos \theta, \;y=r\sin \theta, \; dxdy=rdrd\theta$

区域在x轴以上的条件是:

$0\leq\theta\leq\pi$

区域在半径为2的圆内的条件是

$0\leq r \leq 2$

有了这些条件和转换，积分变成:

$\int^\pi_0 \int^2_0(r^2\cos^2(\theta)+r^2\sin^2(\theta))rdrd\theta=\int^\pi_0 \int^2_0r^3drd\theta$

$=\int^\pi_0 (\frac{1}{4}r^4|^2_0)d\theta=\int^\pi_0 4d\theta=4\theta|^\pi_0=4\pi$

报告一个错误

示例问题3:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(10cos(x^{2} + y^{2}))dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.49\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.063,0.998)\text{ and }\overrightarrow{u_2}=(-0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

9.51

1.58

-28.53

2.38

正确答案:

9.51

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(10cos(x^{2} + y^{2}))dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}10\cdot rcos(r^{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.998}{0.063})=0.48\pi;\theta_2=arctan(\frac{-0.685}{-0.729})=1.24\pi\\&\text{Now, utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{0.48\pi}^{1.24\pi}\int_{0}^{1.49}10\cdot rcos(r^{2}))drd\theta=(5sin(r^{2}))d\theta|_{0}^{1.49}\\&\int_{0.48\pi}^{1.24\pi}(3.983)d\theta=(3.983\cdot \theta)|_{0.48\pi}^{1.24\pi}=9.51\end{align*}$

报告一个错误

示例问题4:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(25e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{2})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.960,0.279)\text{ and }\overrightarrow{u_2}=(-0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-17.38

-8.69

-2.9

69.52

正确答案:

-17.38

解释：

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

$\begin{align*}&\theta_1=arctan(\frac{0.279}{0.960})=0.09\pi;\theta_2=arctan(\frac{-0.976}{-0.218})=1.43\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.09\pi}^{1.43\pi}\int_{0}^{1.77}-\frac{(25\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{2})drd\theta=(\frac{(25e^{(-\frac{(3\cdot r^{2})}{2})})}{6})d\theta|_{0}^{1.77}\\&\int_{0.09\pi}^{1.43\pi}(-4.129)d\theta=(-4.129\cdot \theta)|_{0.09\pi}^{1.43\pi}=-17.38\end{align*}$

报告一个错误

示例问题5:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(43cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.397,0.918)\text{ and }\overrightarrow{u_2}=(-0.951,-0.309)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

563.3

93.88

-469.42

18.78

正确答案:

93.88

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(43cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}43\cdot rcos(\frac{r^{2}}{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.918}{0.397})=0.37\pi;\theta_2=arctan(\frac{-0.309}{-0.951})=1.1\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}\\&\int_{0.37\pi}^{1.1\pi}\int_{0}^{1.94}43\cdot rcos(\frac{r^{2}}{2}))drd\theta=(43sin(\frac{r^{2}}{2}))d\theta|_{0}^{1.94}\\&\int_{0.37\pi}^{1.1\pi}(40.94)d\theta=(40.94\cdot \theta)|_{0.37\pi}^{1.1\pi}=93.88\end{align*}$

报告一个错误

示例问题6:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-41\cdot (\frac{3}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.73\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-58.17

-72.71

1745.13

-290.85

正确答案:

-290.85

解释：

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

$\begin{align*}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{e^{ar^2}}{2aln(b)}\\&\int_{0.34\pi}^{1.32\pi}\int_{0}^{1.73}-41\cdot (\frac{3}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)drd\theta=(-\frac{(123\cdot 3^{(\frac{(2\cdot r^{2})}{3})})}{(4\cdot 2^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{3}{2}))})d\theta|_{0}^{1.73}\\&\int_{0.34\pi}^{1.32\pi}(-94.47)d\theta=(-94.47\cdot \theta)|_{0.34\pi}^{1.32\pi}=-290.85\end{align*}$

报告一个错误

示例问题7:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-50e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.562,-0.827)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-1088.88

-45.37

-272.22

1088.88

正确答案:

-272.22

解释：

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-50e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-50\cdot re^{(\frac{(2\cdot r^{2})}{3})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{0.750})=0.23\pi;\theta_2=arctan(\frac{-0.827}{-0.562})=1.31\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.23\pi}^{1.31\pi}\int_{0}^{1.31}(-50\cdot re^{(\frac{(2\cdot r^{2})}{3})})drd\theta=(-\frac{(75e^{(\frac{(2\cdot r^{2})}{3})})}{2})d\theta|_{0}^{1.31}\\&\int_{0.23\pi}^{1.31\pi}(-80.23)d\theta=(-80.23\cdot \theta)|_{0.23\pi}^{1.31\pi}=-272.22\end{align*}$

报告一个错误

示例问题8:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(20cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.5\text{ and }1.35\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.876,0.482)\text{ and }\overrightarrow{u_2}=(-0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

11.09

-44.35

4.44

22.18

正确答案:

22.18

解释：

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

$\begin{align*}&\theta_1=arctan(\frac{0.482}{-0.876})=0.84\pi;\theta_2=arctan(\frac{-0.988}{-0.156})=1.45\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}\\&\int_{0.84\pi}^{1.45\pi}\int_{0.5}^{1.35}(20\cdot rcos(\frac{(2\cdot r^{2})}{3}))drd\theta=(15sin(\frac{(2\cdot r^{2})}{3}))d\theta|_{0.5}^{1.35}\\&\int_{0.84\pi}^{1.45\pi}(11.57)d\theta=(11.57\cdot \theta)|_{0.84\pi}^{1.45\pi}=22.18\end{align*}$

报告一个错误

示例问题9:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(16sin(x^{2} + y^{2}))dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.33\text{ and }1.76\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.982,0.187)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

-93.17

7.76

15.53

46.59

正确答案:

46.59

解释：

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

$\begin{align*}&\theta_1=arctan(\frac{0.187}{-0.982})=0.94\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.94\pi}^{1.87\pi}\int_{0.33}^{1.76}(16\cdot rsin(r^{2}))drd\theta=(-8cos(r^{2}))d\theta|_{0.33}^{1.76}\\&\int_{0.94\pi}^{1.87\pi}(15.94)d\theta=(15.94\cdot \theta)|_{0.94\pi}^{1.87\pi}=46.59\end{align*}$

报告一个错误

示例问题10:极坐标下的二重积分

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{49}{(\frac{2}{3})^{(x^{2}+ y^{2})}})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.16\text{ and }0.97\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

可能的答案:

20.54

123.25

616.27

-493.02

正确答案:

123.25

解释：

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

$\begin{align*}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.34\pi}^{1.77\pi}\int_{0.16}^{0.97}(\frac{(49\cdot r)}{(\frac{2}{3})^{(r^{2})}})drd\theta=(-\frac{(49\cdot (\frac{3}{2})^{(r^{2})})}{(2ln(\frac{2}{3}))})d\theta|_{0.16}^{0.97}\\&\int_{0.34\pi}^{1.77\pi}(27.44)d\theta=(27.44\cdot \theta)|_{0.34\pi}^{1.77\pi}=123.25\end{align*}$

报告一个错误

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雅罗斯瓦夫
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