Transformation Between Line Integral and Surface Integral
Let \(\vec{F}\) be a vector field with continuous first-order partial derivatives in a domain containing an open surface \(S\) bounded by a simple closed curve \(C\). Then:
\[ \oint\limits_{C} \vec{F} \cdot d\vec{r} = \iint\limits_{S} (\nabla \times \vec{F}) \cdot \hat{n} \, dS \]where:
- \(\hat{n}\) is the unit normal vector to surface \(S\)
- \(C\) is traversed in the positive direction (right-hand rule)
Visual Interpretation
Stokes' Theorem establishes a fundamental relationship between:
Line Integral
Circulation of \(\vec{F}\) around boundary curve \(C\)
\[ \oint\limits_{C} \vec{F} \cdot d\vec{r} \]Surface Integral
Flux of curl \(\vec{F}\) through surface \(S\)
\[ \iint\limits_{S} (\nabla \times \vec{F}) \cdot \hat{n} \, dS \]Key Points
- The surface \(S\) must be orientable (two-sided)
- The curve \(C\) must be piecewise smooth
- The positive direction of \(C\) follows the right-hand rule with respect to \(\hat{n}\)
- Special cases:
- When \(\nabla \times \vec{F} = \vec{0}\), the field is conservative and circulation is zero
- For a planar surface in xy-plane, reduces to Green's Theorem
Physical Interpretation
Stokes' Theorem has profound physical meanings in:
- Fluid dynamics: Relates macroscopic circulation to microscopic rotation
- Electromagnetism: Connects line integrals of electric fields to magnetic flux (Faraday's Law)
- Vortex theory: Shows how vorticity in fluids generates circulation
Computational Aspects
In practice, Stokes' Theorem is used to:
- Convert difficult line integrals into easier surface integrals
- Calculate circulation around complex curves
- Verify conservative vector fields
- Solve problems in electromagnetism and fluid mechanics
Example 1: Verify Stokes' theorem for \(\oint\limits_{C} (x^2 + y^2)dx - 2xy dy\) around the rectangle bounded by \(x = \pm a\), \(y = 0\), and \(y = b\).
Solution:Part 1: Using Stokes' Theorem (Surface Integral)
Given vector field:
\[ \vec{F} = (x^2 + y^2)\hat{i} - 2xy\hat{j} \]Compute the curl:
\[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 + y^2 & -2xy & 0 \end{vmatrix} = -4y\hat{k} \]For the rectangular surface in xy-plane, the unit normal is \(\hat{n} = \hat{k}\):
\[ (\nabla \times \vec{F}) \cdot \hat{n} = -4y \]Compute the surface integral:
\[ \iint\limits_{S} (\nabla \times \vec{F}) \cdot \hat{n} \, dS = -4 \iint\limits_{R} y \, dx dy \] \[ = -4 \int_{x=-a}^{a} \int_{y=0}^{b} y \, dy dx = -4 \int_{-a}^{a} \left[ \frac{y^2}{2} \right]_0^b dx \] \[ = -4 \cdot \frac{b^2}{2} \int_{-a}^{a} dx = -2b^2 [x]_{-a}^a = -4ab^2 \]Part 2: Direct Line Integral Verification
The boundary \(C\) consists of four segments:
Segment 1: Bottom edge (y=0 from x=-a to x=a)
\[ y = 0, dy = 0 \] \[ \int_{-a}^{a} (x^2 + 0)dx - 0 = \left[ \frac{x^3}{3} \right]_{-a}^a = \frac{2a^3}{3} \]Segment 2: Right edge (x=a from y=0 to y=b)
\[ x = a, dx = 0 \] \[ \int_{0}^{b} 0 - 2a y dy = -a b^2 \]Segment 3: Top edge (y=b from x=a to x=-a)
\[ y = b, dy = 0 \] \[ \int_{a}^{-a} (x^2 + b^2)dx = -\frac{2a^3}{3} - 2ab^2 \]Segment 4: Left edge (x=-a from y=b to y=0)
\[ x = -a, dx = 0 \] \[ \int_{b}^{0} 0 - 2(-a)y dy = -a b^2 \]Summing all segments:
\[ \frac{2a^3}{3} - a b^2 - \frac{2a^3}{3} - 2a b^2 - a b^2 = -4a b^2 \]Verification Complete
\[ \text{Both methods yield: } \boxed{ -4ab^2 } \]Stokes' theorem is verified for this case.
Note on Orientation
The negative sign indicates the clockwise circulation when viewing from the positive z-axis. The consistent result from both methods confirms the proper application of Stokes' Theorem, including the correct orientation of the boundary curve.
Example 2: Evaluate \(\oint\limits_{C} (-y^3)dx + x^3 dy\) where \(C\) is the boundary of the region \(x^2 + y^2 \leq 1\) in the plane \(z = 0\)..
Solution: \[ \vec{F} = -y^3 \hat{i} + x^3 \hat{j} \] \[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -y^3 & x^3 & 0 \end{vmatrix} = (0-0)\hat{i} - (0-0)\hat{j} + (3x^2 + 3y^2)\hat{k} = 3(x^2 + y^2)\hat{k} \] \[ \oint\limits_{C} \vec{F} \cdot d\vec{r} = \iint\limits_{S} (\nabla \times \vec{F}) \cdot \hat{n} \, dS \]For the surface \(z = 0\) (xy-plane), the unit normal is \(\hat{n} = \hat{k}\):
\[ (\nabla \times \vec{F}) \cdot \hat{n} = 3(x^2 + y^2) \]The region is a unit disk, so we use polar coordinates:
\[ x = r\cos\theta, \quad y = r\sin\theta, \quad dS = r dr d\theta \] \[ x^2 + y^2 = r^2 \]Integration limits:
\[ r \in [0,1], \quad \theta \in [0,2\pi] \] \[ \iint\limits_{S} 3(x^2 + y^2) dS = 3 \int_{0}^{2\pi} \int_{0}^{1} r^2 \cdot r dr d\theta \] \[ = 3 \int_{0}^{2\pi} d\theta \int_{0}^{1} r^3 dr \] \[ = 3 \cdot 2\pi \cdot \left[ \frac{r^4}{4} \right]_0^1 \] \[ = 6\pi \cdot \frac{1}{4} = \frac{3\pi}{2} \] \[ \boxed{\dfrac{3\pi}{2}} \]Example 3: Evaluate \(\oint\limits_{C} (y dx + z dy + x dz)\) where \(C\) is the curve of intersection of the sphere \(x^2 + y^2 + z^2 = a^2\) and the plane \(x + z = a\).
Solution: \[ \vec{F} = y \hat{i} + z \hat{j} + x \hat{k} \] \[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & z & x \end{vmatrix} = -\hat{i} - \hat{j} - \hat{k} \]The curve \(C\) is the intersection of the sphere and plane. The surface \(S\) we'll use is the portion of the plane \(x + z = a\) inside the sphere.
Normal vector to the plane:
\[ \vec{n} = \frac{1}{\sqrt{2}}(1, 0, 1) \quad \text{(unit normal)} \]Projection area in xy-plane: substitute \(z = a - x\) into sphere equation:
\[ x^2 + y^2 + (a - x)^2 = a^2 \Rightarrow 2x^2 - 2ax + y^2 = 0 \] \[ \Rightarrow \frac{(x-a/2)^2}{(a/2)^2} + \frac{y^2}{(a/\sqrt{2})^2} = 1 \quad \text{(ellipse)} \] \[ \oint\limits_{C} \vec{F} \cdot d\vec{r} = \iint\limits_{S} (\nabla \times \vec{F}) \cdot \hat{n} \, dS \] \[ = \iint\limits_{S} (-\hat{i} - \hat{j} - \hat{k}) \cdot \left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}\right) dS \] \[ = \iint\limits_{S} \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) dS = -\sqrt{2} \iint\limits_{S} dS \]The area of the elliptical surface \(S\):
\[ \text{Area} = \pi \left(\frac{a}{2}\right) \left(\frac{a}{\sqrt{2}}\right) = \frac{\pi a^2}{2\sqrt{2}} \] \[ \oint\limits_{C} \vec{F} \cdot d\vec{r} = -\sqrt{2} \times \frac{\pi a^2}{2\sqrt{2}} = -\frac{\pi a^2}{2} \]Example 4: Evaluate \(\oint\limits_{C} \vec{F} \cdot d\vec{r}\) for \(\vec{F} = (y-z+2)\hat{i} + (yz+4)\hat{j} - xz\hat{k}\) where \(C\) is the boundary of the cube \(x=0,y=0,z=0,x=2,y=2,z=2\) above the xy-plane.
Solution: \[ \vec{F} = (y-z+2)\hat{i} + (yz+4)\hat{j} - xz\hat{k} \] \[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y-z+2 & yz+4 & -xz \end{vmatrix} \] \[ = \left(\frac{\partial(-xz)}{\partial y} - \frac{\partial(yz+4)}{\partial z}\right)\hat{i} - \left(\frac{\partial(-xz)}{\partial x} - \frac{\partial(y-z+2)}{\partial z}\right)\hat{j} \] \[ + \left(\frac{\partial(yz+4)}{\partial x} - \frac{\partial(y-z+2)}{\partial y}\right)\hat{k} \] \[ = (0 - y)\hat{i} - (-z - (-1))\hat{j} + (0 - 1)\hat{k} \] \[ = -y\hat{i} + (z-1)\hat{j} - \hat{k} \] \[ \oint\limits_{C} \vec{F} \cdot d\vec{r} = \iint\limits_{S} (\nabla \times \vec{F}) \cdot \hat{n} \, dS \]For the top face of the cube (\(z=2\)):
\[ \hat{n} = \hat{k}, \quad dS = dx dy \] \[ (\nabla \times \vec{F}) \cdot \hat{n} = -1 \] \[ \iint\limits_{S_{\text{top}}} (\nabla \times \vec{F}) \cdot \hat{n} \, dS = -1 \times \text{Area} = -4 \]For the other faces (x=0, x=2, y=0, y=2):
\[ \text{The dot products with their normals evaluate to 0} \] \[ \oint\limits_{C} \vec{F} \cdot d\vec{r} = -4 \]Example 5:Evaluate \(\oint\limits_{C} (2x-y)dx - yz^2 dy - y^2 z dz\) over the upper half surface of the sphere \(x^2 + y^2 + z^2 = 1\) bounded by the xy-plane.
Solution: \[ \vec{F} = (2x-y)\hat{i} - yz^2\hat{j} - y^2z\hat{k} \] \[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2x-y & -yz^2 & -y^2z \end{vmatrix} \] \[ = \left(\frac{\partial(-y^2z)}{\partial y} - \frac{\partial(-yz^2)}{\partial z}\right)\hat{i} - \left(\frac{\partial(-y^2z)}{\partial x} - \frac{\partial(2x-y)}{\partial z}\right)\hat{j} \] \[ + \left(\frac{\partial(-yz^2)}{\partial x} - \frac{\partial(2x-y)}{\partial y}\right)\hat{k} \] \[ = (-2yz + 2yz)\hat{i} - (0 - 0)\hat{j} + (0 - (-1))\hat{k} \] \[ = \hat{k} \] \[ \oint\limits_{C} \vec{F} \cdot d\vec{r} = \iint\limits_{S} (\nabla \times \vec{F}) \cdot \hat{n} \, dS \]For the upper hemisphere (\(z \geq 0\)):
\[ \hat{n} = x\hat{i} + y\hat{j} + z\hat{k} \quad \text{(outward normal)} \] \[ (\nabla \times \vec{F}) \cdot \hat{n} = \hat{k} \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = z \]Using spherical coordinates for the upper hemisphere:
\[ x = \sin\theta\cos\phi, \quad y = \sin\theta\sin\phi, \quad z = \cos\theta \] \[ dS = \sin\theta d\theta d\phi \] \[ \iint\limits_{S} z \, dS = \int_{0}^{2\pi} \int_{0}^{\pi/2} \cos\theta \cdot \sin\theta \, d\theta d\phi \] \[ = 2\pi \int_{0}^{\pi/2} \frac{\sin 2\theta}{2} d\theta = \pi \left[-\frac{\cos 2\theta}{2}\right]_0^{\pi/2} \] \[ = \pi \left( \frac{1}{2} + \frac{1}{2} \right) = \pi \]