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曲线积分计算

बनाया गया: 1 फ़रवरी 2025

का उपयोग करके उत्तर दिया गया O3 Mini High द्वारा Chat01

प्रश्न

\begin{aligned}&20、(12分)已知向曲线L是球面x^2+y^2+z^2=2x\text{ 与平面}&2x-z-1=0\text{ 的交线,从}z\text{轴正向往}z\text{ 轴负向看为逆时针方}&\text{向,计算曲线积分}&I=\int_L\left(6xyz-yz^2\right)\mathrm{d}x+2x^2z\mathrm{d}y+xyz\mathrm{d}z .\end{aligned}

विचार

Analyzing intersections

I’m piecing together the intersection of a sphere and a plane. The sphere, centered at (1,0,0) with radius 1, intersects the plane z=2x−1z = 2x - 1. This forms a circle, and the orientation is counterclockwise.

Leveraging Stokes' theorem

OK, let me see. I'm exploring Stokes' theorem for the closed curve L, considering curl F\text{curl} \, F and a surface bounded by L, typically the sphere-plane intersection.

Mapping the curl

I’m calculating curl F\text{curl} \, F by taking partial derivatives of P,Q,P, Q, and RR. This involves complex derivatives like ∂R/∂y\partial R / \partial y and ∂Q/∂z\partial Q / \partial z.

Determining the orientation

I’ve found curl F\text{curl} \, F to be (xz−2x2,6xy−3yz,−2xz+z2)(x z - 2 x^2, 6xy - 3yz, -2xz + z^2). Next, I’ll evaluate the surface integral over the plane inside the sphere.

Confirming orientation

I’m ensuring the unit normal of the plane aligns with the counterclockwise orientation of curve L, consistent with the given direction from the positive z-axis to the negative z-axis.

Comparing vectors

I’m noting that vector nn should align with the counterclockwise movement of the curve when viewed from the positive z-axis, using the right-hand rule to determine the correct orientation.

Aligning plane parameters

Mapping out the plane's equation, I’m ensuring the orientation aligns with the counterclockwise view from the positive z-axis, maintaining the boundary alignment.

Aligning orientation

I’m tweaking the normal vector nn on the plane to ensure the boundary curve LL is counterclockwise from the positive z-axis, using the right-hand rule.

Mapping the process

I’m opting for Stokes' theorem to streamline the surface integral. By parameterizing the curve and using the plane's normal vector, I can simplify calculations with curl FF and nn.

Confirming orientation

I’m mapping out how the computed normal vector aligns with our orientation from the positive z-axis to the negative z-axis, ensuring it correctly matches the counterclockwise rotation.

Choosing the right vector

I’m suggesting the normal vector n=(−2,0,1)/5n = (-2, 0, 1)/\sqrt{5} so that when viewed from the positive z-axis towards the negative z-axis, the curve appears counterclockwise.

Determining the normal vector

I’m finalizing the normal vector n=(−2,0,1)/5n = (-2, 0, 1)/\sqrt{5} based on its positive z-component. This choice simplifies the flux calculation by substituting z=2x−1z = 2x - 1 into curl F\text{curl } F.

Substituting and simplifying

Piecing together the components of curl F\text{curl } F by substituting z=2x−1z = 2x - 1 into (x,y,z)(x, y, z), I simplify to (−x,3y,−2x+1)(-x, 3y, -2x + 1).

Completing the square

I’m working through completing the square for the xx and yy terms in the equation 5x2−6x+y2+1=05x^2 - 6x + y^2 + 1 = 0 to identify the disk's boundary.

Adjusting to inclined planes

I’m noting that for inclined planes like S:z=2x−1S: z = 2x - 1, the surface element dSdS incorporates the inclination, leading to 5 dx dy\sqrt{5} \, dx \, dy.

Confirming integrand accuracy

I’m verifying the integration by recalculating (curlF)⋅n(curl F) \cdot n with n=(−2,0,1)/5n = (-2, 0, 1)/\sqrt{5}. The simplified integrand consistently yields 1.

Re-checking geometrical intersections

I'm piecing together the intersection of the plane z=2x−1z = 2x - 1 with the sphere, ensuring the projection onto the xx-yy plane is correctly defined.

Verifying calculations

I’m ensuring the sphere equation x2+y2+z2=2xx^2 + y^2 + z^2 = 2x is accurately substituted with z=2x−1z = 2x - 1, confirming xx and yy are real.

Analyzing intersection

OK, let's see. The sphere, centered at (1,0,0)(1,0,0) with radius 1, intersects the inclined plane z=2x−1z = 2x - 1. The circular intersection's radius measures 2/52/\sqrt{5}.

Revisiting sphere-plane intersection

I’m recalculating the intersection of the sphere (x−1)2+y2+(2x−1)2=1(x-1)^2 + y^2 + (2x-1)^2 = 1 with the plane z=2x−1z = 2x - 1, ensuring accuracy in the xx-yy plane projection.

Reevaluating sphere-plane intersection

OK, let me see. Substituting z=2x−1z = 2x - 1 in the sphere equation simplifies to 5x2−6x+y2=−15x^2 - 6x + y^2 = -1, a quadratic likely invalid for this intersection.

Rechecking intersections

I’m noting the plane intersects the sphere in a circle with a radius of 4/5\sqrt{4/5}. The plane is tangent, and the intersection should form a circle.

Reassessing sphere-plane intersection

I’m piecing together the sphere equation (x−1)2+y2+z2=1(x-1)^2 + y^2 + z^2 = 1 and the plane z=2x−1z = 2x - 1, finding the intersection might be a degenerate circle.

Reworking the equation

I’m carefully checking the equation 5(x−3/5)2+y2=4/55(x - 3/5)^2 + y^2 = 4/5, ensuring the semi-axis lengths are 4/25\sqrt{4/25} in the x-direction and 4/5\sqrt{4/5} in the y-direction.

Calculating intersection dimensions

I’m realizing that the sphere-plane intersection, despite being an ellipse in projection, forms a circle in 3D space, with defined semi-axes of 2/5 and 2/√5.

Mapping domain area

I’m calculating the area of projection in the x-y plane for region D, finding the domain area is 4π55\frac{4\pi}{5\sqrt{5}}.

Confirming orientation

I’m verifying that the normal vector nn aligns with the boundary's orientation through the right-hand rule. The boundary in the x-y plane exhibits a counterclockwise orientation.

Confirming orientation logic

I’m aligning the normal vector nn with the counterclockwise boundary orientation from the positive z-axis, ensuring it aligns with the expected positive flux direction.

Reassessing orientation

I’m confirming that the normal vector nn aligns with the counterclockwise boundary orientation from the positive z-axis, ensuring it matches the expected positive flux direction.

Determining area

I’m mapping out the area of ellipse DD in the x-y plane, first by deriving its equation from the sphere-plane intersections, then calculating the area using the semi-major and semi-minor axes.

Confirming calculations

I’m validating the area of the ellipse DD in the x-y plane as 4π55\frac{4\pi}{5\sqrt{5}}, ensuring all integration steps and the dot product computation are accurate.

Breaking down calculations

I’m working through the curl of FF on SS, found as (−x,3y,−2x+1)(-x, 3y, -2x+1), and noting the upward-oriented unit normal vector nn as (−2,0,1)/5(-2, 0, 1)/\sqrt{5}.

Calculating flux

I’m noting that the flux through DD is 1, and the area of the ellipse DD in the x-y plane is 4π55\frac{4\pi}{5\sqrt{5}}.

Thought about curve integral calculation for 1m 14s

उत्तर

我们要求计算曲线积分

I=∫L[(6xyz−yz2) dx+2x2z dy+xyz dz],I=\int_L\Big[(6xyz-yz^2)\,\mathrm{d}x+2x^2z\,\mathrm{d}y+xyz\,\mathrm{d}z\Big],

其中 LL 是球面

x2+y2+z2=2xx^2+y^2+z^2=2x

与平面

2x−z−1=02x-z-1=0

的交线，并且规定：从 zz 轴正向往 zz 轴负向看时，LL 的正向为逆时针方向。

下面给出一种利用斯托克斯公式求解的详细步骤。

1. 将曲线 LL 的几何意义写清

首先，球面

x2+y2+z2=2xx^2+y^2+z^2=2x

可以改写为

x2−2x+y2+z2=0⟹(x−1)2+y2+z2=1.x^2-2x+y^2+z^2=0\quad\Longrightarrow\quad (x-1)^2+y^2+z^2=1.

因此，球心为 (1,0,0)(1,0,0)，半径为 11。

平面

2x−z−1=0⟹z=2x−1.2x-z-1=0\quad\Longrightarrow\quad z=2x-1.

故 LL 为球与平面的交线，必为一圆。

2. 用斯托克斯公式化简曲线积分

记向量场

F(x,y,z)=(P,Q,R)=(6xyz−yz2,  2x2z,  xyz).\mathbf{F}(x,y,z)=(P,Q,R)=(6xyz-yz^2,\;2x^2z,\; xyz).

斯托克斯公式告诉我们

∮LF⋅dr=∬S(∇×F)⋅n dS,\oint_{L}\mathbf{F}\cdot\mathrm{d}\mathbf{r}=\iint_{S}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,\mathrm{d}S,

其中 SS 为 LL 的任一定向光滑有界曲面，n\mathbf{n} 是 SS 上与 LL 一致的单位正法向。由于 LL 是球面与平面的交线，选取平面 2x−z−1=02x-z-1=0 所围成的区域作为 SS 是比较方便的。

首先我们求 ∇×F\nabla\times\mathbf{F}：

  • 记 P=6xyz−yz2,Q=2x2z,R=xyz.P=6xyz-yz^2,\quad Q=2x^2z,\quad R=xyz.
  • 按照叉乘公式 ∇×F=(∂R∂y−∂Q∂z,  ∂P∂z−∂R∂x,  ∂Q∂x−∂P∂y),\nabla\times\mathbf{F}=\Big(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z},\;\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x},\;\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big), 分别计算：

    • ∂R∂y=∂(xyz)∂y=xz\frac{\partial R}{\partial y}=\frac{\partial (xyz)}{\partial y}=xz,

      ∂Q∂z=∂(2x2z)∂z=2x2\frac{\partial Q}{\partial z}=\frac{\partial (2x^2z)}{\partial z}=2x^2,

      故第一分量为

      xz−2x2=x(z−2x).xz-2x^2=x(z-2x).

    • ∂P∂z=∂(6xyz−yz2)∂z=6xy−2yz\frac{\partial P}{\partial z}=\frac{\partial (6xyz-yz^2)}{\partial z}=6xy-2yz,

      ∂R∂x=∂(xyz)∂x=yz\frac{\partial R}{\partial x}=\frac{\partial (xyz)}{\partial x}=yz,

      故第二分量为

      6xy−2yz−yz=6xy−3yz.6xy-2yz-yz=6xy-3yz.

    • ∂Q∂x=∂(2x2z)∂x=4xz\frac{\partial Q}{\partial x}=\frac{\partial (2x^2z)}{\partial x}=4xz,

      ∂P∂y=∂(6xyz−yz2)∂y=6xz−z2\frac{\partial P}{\partial y}=\frac{\partial (6xyz-yz^2)}{\partial y}=6xz-z^2,

      故第三分量为

      4xz−(6xz−z2)=−2xz+z2.4xz-(6xz-z^2)=-2xz+z^2.

因此，

∇×F=(xz−2x2,  6xy−3yz,  −2xz+z2).\nabla\times\mathbf{F}=(xz-2x^2,\;6xy-3yz,\;-2xz+z^2).

3. 将积分曲面 SS 选在平面上，并确定正向

我们选取的曲面 SS 就是平面 2x−z−1=02x-z-1=0 内被 LL 所围成的区域。为了利用 z=f(x,y)z=f(x,y) 的形式，我们写出平面方程

z=2x−1.z=2x-1.

当曲面以图形 z=f(x,y)z=f(x,y) 表示时，其“向上”的单位正法向为

n=(−fx,−fy,1)1+fx2+fy2.\mathbf{n}=\frac{(-f_x,-f_y,1)}{\sqrt{1+f_x^2+f_y^2}}.

这里 fx=∂(2x−1)∂x=2,  fy=0f_x=\frac{\partial (2x-1)}{\partial x}=2,\; f_y=0，故

n=(−2,0,1)1+4=(−2,0,1)5.\mathbf{n}=\frac{(-2,0,1)}{\sqrt{1+4}}=\frac{(-2,0,1)}{\sqrt5}.

注意：此 n\mathbf{n} 的 zz 分量为正，因此按右手定则，曲线 LL 的正向（即 SS 所诱导的边界方向）为“自上而下看”时的逆时针方向，与题中“从 zz 轴正向往 zz 轴负向看为逆时针方向”是一致的。

4. 在 SS 上计算 (∇×F)⋅n(\nabla\times\mathbf{F})\cdot\mathbf{n}

因为在 SS 上有 z=2x−1z=2x-1，我们将 ∇×F\nabla\times\mathbf{F} 中的 zz 换成 2x−12x-1：

  • 第一分量： xz−2x2=x(2x−1)−2x2=2x2−x−2x2=−x.xz-2x^2=x(2x-1)-2x^2=2x^2-x-2x^2=-x.
  • 第二分量： 6xy−3yz=6xy−3y(2x−1)=6xy−6xy+3y=3y.6xy-3yz=6xy-3y(2x-1)=6xy-6xy+3y=3y.
  • 第三分量： −2xz+z2=−2x(2x−1)+(2x−1)2.-2xz+z^2=-2x(2x-1)+(2x-1)^2. 计算： −2x(2x−1)=−4x2+2x,(2x−1)2=4x2−4x+1,-2x(2x-1)=-4x^2+2x,\quad (2x-1)^2=4x^2-4x+1, 故第三分量为 (−4x2+2x)+(4x2−4x+1)=−2x+1.(-4x^2+2x)+(4x^2-4x+1)=-2x+1.

因此，在 SS 上有

∇×F=(−x, 3y, −2x+1).\nabla\times\mathbf{F} = (-x,\,3y,\,-2x+1).

与单位法向 n=(−2,0,1)5\mathbf{n}=\frac{(-2,0,1)}{\sqrt5} 点乘：

(∇×F)⋅n=(−x, 3y, −2x+1)⋅(−2,0,1)5=2x+(−2x+1)5=15.(\nabla\times\mathbf{F})\cdot\mathbf{n} = (-x,\,3y,\,-2x+1)\cdot\frac{(-2,0,1)}{\sqrt5} = \frac{2x+(-2x+1)}{\sqrt5}=\frac{1}{\sqrt5}.

另一方面，由于平面给出的图像 z=2x−1z=2x-1 的面积元为

dS=1+fx2+fy2 dx dy=1+4 dx dy=5 dx dy,\mathrm{d}S=\sqrt{1+f_x^2+f_y^2}\,\mathrm{d}x\,\mathrm{d}y=\sqrt{1+4}\,\mathrm{d}x\,\mathrm{d}y=\sqrt5\,\mathrm{d}x\,\mathrm{d}y,

所以积分的被积函数为

(∇×F)⋅n dS=15⋅5 dx dy=dx dy.(\nabla\times\mathbf{F})\cdot\mathbf{n}\,\mathrm{d}S=\frac{1}{\sqrt5}\cdot\sqrt5\,\mathrm{d}x\,\mathrm{d}y =\mathrm{d}x\,\mathrm{d}y.

综上，

I=∮LF⋅dr=∬S(∇×F)⋅n dS=∬D1 dx dy,I=\oint_{L}\mathbf{F}\cdot\mathrm{d}\mathbf{r}=\iint_{S}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,\mathrm{d}S =\iint_D 1\,\mathrm{d}x\,\mathrm{d}y,

其中 DD 是 SS 在 xyxy 平面上的投影区域。

5. 求区域 DD 的面积

区域 DD 是由 LL 在 xyxy 平面上的投影。将平面 z=2x−1z=2x-1 的表达代入球面方程

(x−1)2+y2+z2=1,(x-1)^2+y^2+z^2=1,

得

(x−1)2+y2+(2x−1)2=1.(x-1)^2+y^2+(2x-1)^2=1.

计算各项：

  • (x−1)2=x2−2x+1(x-1)^2=x^2-2x+1,
  • (2x−1)2=4x2−4x+1(2x-1)^2=4x^2-4x+1.

因此，

x2−2x+1+4x2−4x+1+y2=5x2−6x+y2+2=1,x^2-2x+1+4x^2-4x+1+y^2=5x^2-6x+y^2+2=1,

整理得

5x2−6x+y2=−1.5x^2-6x+y^2= -1.

为便于讨论，将 xx 的平方项配方：

5x2−6x=5(x2−65x)=5[(x−35)2−925].5x^2-6x=5\Bigl(x^2-\frac{6}{5}x\Bigr)=5\Bigl[\Bigl(x-\frac{3}{5}\Bigr)^2-\frac{9}{25}\Bigr].

代入上式，

5[(x−35)2−925]+y2=−1,5\Bigl[\Bigl(x-\frac{3}{5}\Bigr)^2-\frac{9}{25}\Bigr]+y^2=-1,

即

5(x−35)2+y2=95−1=45.5\Bigl(x-\frac{3}{5}\Bigr)^2+y^2=\frac{9}{5}-1=\frac{4}{5}.

写成标准形式，

(x−35)2425+y245=1.\frac{\Bigl(x-\frac{3}{5}\Bigr)^2}{\frac{4}{25}}+\frac{y^2}{\frac{4}{5}}=1.

因此，DD 在 xyxy 平面上是一个椭圆，其半轴长为

a=425=25,b=45=25.a=\sqrt{\frac{4}{25}}=\frac{2}{5},\quad b=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt5}.

椭圆的面积为

Area(D)=πab=π⋅25⋅25=4π55.\text{Area}(D)=\pi a b=\pi\cdot\frac{2}{5}\cdot\frac{2}{\sqrt5}=\frac{4\pi}{5\sqrt5}.

6. 得出最终结果

因此，

I=∬Ddx dy=Area(D)=4π55.I=\iint_D\mathrm{d}x\,\mathrm{d}y=\text{Area}(D)=\frac{4\pi}{5\sqrt5}.

最终答案

曲线积分的值为

4π55.\boxed{\frac{4\pi}{5\sqrt5}}.

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