# 4.8 重积分

## 4.8.1 重积分的计算

在Integrate()函数中嵌套多层Integrate()函数可以对重积分进行计算。

例：计算积分$$\int\_{0}^{2} dx \int\_{x}^{2} e^{-y^2} dy$$的值。

```python
[]:integrate(integrate(exp(-y\*\*2), (y, x, 2)), (x, 0, 2)).simplify()
```

\[]:![](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-0129ef8233b4a66d29dfac0f6dd9707c4e8029a9%2Fa72a0415c477146d1e4571d484d5c41f.png?alt=media)

例：计算三重积分$$\iiint \limits\_{\Omega} z dxdydz$$其中为平面与三个坐标面，，围成的闭区域。（）

```python
[]:plot3d(1 - x - y, (x, 0, 1), (y, 0, 1), aspect_ratio=(1, 1, 1))
```

![C:\Users\Johan\AppData\Local\Microsoft\Windows\INetCache\Content.MSO\E18883DA.tmp](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-58b36b107a97e444332bdfa3262b466badacfdc5%2Fde1daf9cac3636ef5f72539cfc6ff85c.png?alt=media)

```python
[]:<sympy.plotting.plot.Plot at 0x12701ba8\>

[]:integrate(integrate(integrate(z, (z, 0, 1-x-y)), (y, 0, 1-x)), (x, 0, 1))
```

![](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-77ee981c95f131aa3e3b74f9c7c2b9e57f74a540%2Ff465a7e3a9ea5cbbd168b2b71e124b0e.png?alt=media)\[]:

## 4.8.2 重积分的应用

### 求曲面间围成的面积

例：求曲面$$z=x^2+y^2$$和$$z=2-x^2-y^2$$围成的体积

首先使用plot3d绘制出曲面图形:

```python
[]:plot3d((x\*\*2+y\*\*2), (2-x\*\*2-y\*\*2))
```

![C:\Users\Johan\AppData\Local\Microsoft\Windows\INetCache\Content.MSO\8B72AC8.tmp](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-3ffb1a68be3606e3194f03404da3a90ee5f6c2aa%2Fadc35c7f3c9466552aeed4bc3b4f9e36.png?alt=media)

```python
[]:\<sympy.plotting.plot.Plot at 0x141fc6d8\>

[]:integrate(integrate(integrate(1,(z, r\*\*2, 2-r\*\*2)), (r, 0, 1)), (theta,
0, 2\*pi))
```

\[]:![](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-ae0f35686a9209a9aa5de65a3af45063fe127b18%2F6a7fd950ff83c3ad5ef520c561e8065b.png?alt=media)

例：计算曲面$$z=x^2+2y^2$$和$$z=6-2x^2-y^2$$ 所围成体积

```python
[]:plot3d(x\*\*2+2\*y\*\*2,3-2\*\*2-y\*\*2)
```

由图像可知，所求体积$$V=\int \limits\_D (3-2x^2-y^2-(x^2+2y^2) d\sigma$$，其中积分区域$$D={(x, y)|x^2+y^2<2}$$。使用极坐标计算此积分，$$V=\int\_{0}^{2\pi}d\theta \int\_{0}^{\sqrt{2}} (6r - 3r3) dr$$

```python
integrate(integrate(integrate(6\*r-r\*r\*\*3, (r, 0, sqrt(2)), (theta, 0,
2\*pi))
```

例：求$$\iint \limits\_{D} \left ( x^2+y \right ) dxdy$$, 其中$$D$$是由抛物线$$y=x^2$$和$$x=y^2$$所围平面闭区域。

```python
[]:integrate(integrate(x\*\*2+y, (y, x\*\*2, sqrt(x))), (x,0,1))
```

\[]:![](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-f2e7e61cec4ebbb3d24e93bdb185262401e8ffa0%2F946ba3024d82c4d181c853198b09d42c.png?alt=media)

### 曲线积分

例：设L为$$\begin{cases} x & = e^t + 1\ y & = e^t - 1 \end{cases}$$ 从t=0到$$\log 2$$的一段弧，求曲线面积$$\int \limits\_L xdx + ydy$$

```python
[]:from sympy import Curve, line_integrate, E, ln

from sympy.abc import x, y, t

*[]:C = Curve([E\*\*t + 1, E\*\*t - 1], (t, 0, ln(2)))*

*line_integrate(x + y, C, [x, y])*
```

\[]: ![](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-02d81461a9c9d772f6658d649b146df28b67f8db%2F8a773398109992002c18b98933f65972.png?alt=media)

### 曲面积分

例：计算$$\iint \limits\_{\Sigma} x^2dydz + y^2 dzdx + zdxdy$$，其中$$\Sigma$$是旋转抛物面$$z=1-x^2-y^2(z\ge 0$$的上侧。

接下来，我们二重积分化为三重积分。首先需要补面：$$\Sigma': x^2+y^2\le1, z=0$$,与$$\Sigma$$围成封闭区域$$\Sigma$$。然后，使用Gauss公式进行计算。原积分 $$=\iint\_{\sigma'+\sigma} - \iint^{'}*{\Sigma}= \iiint*{\Omega} - \iint^{'}\_{\Sigma'}$$,

其中,$$\int\_{\Omega} \left ( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \right ) dxdydz =\iiint\_{\Omega} \left ( 2x+2y+1 \right ) dxdydz$$, $$\iint^{'}\_{\Sigma} x^2dydz + y^2 dzdx + zdxdy = 0$$ 化三重积分为三次积分：

$$
\iiint\_{\Omega} \left ( 2x+2y+1 \right ) dxdydz = \int\_{0}^{2\pi} \int\_{0}^{1} r dr \int\_{0}^{1-r^2} (2r \cos \theta + 2r \sin \theta + 1) dz
$$

在Jupyter Lab中对该积分进行计算：

```python
*[]:integrate(integrate(integrate((2\*r\*cos(theta)+2\*r\*sin(theta)+1)\*r*

*, (z3,0, 1-r\*\*2)), (r, 0, 1)), (theta, 0, 2\*pi))*

```

\[]: ![](https://3607972777-files.gitbook.io/~/files/v0/b/gitbook-x-prod.appspot.com/o/spaces%2F-Mjx-9CYfSlrNPw45CMh%2Fuploads%2Fgit-blob-d8460ad111d2ebf2ad9c9c42d80f502601aa6c0f%2Fc7130ba351a79f9287ab1e013ced87c5.png?alt=media)


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