Dated: 30-10-2024
Vector Field
Conservative Vector Fields
The value of integral \(\int_c \vec F \cdot d \vec r\) between 2 points depends on particular paths of integration.
However if this value in independent of the path for a vector field \(\vec F\) then \(\vec F\) is called a conservative vector field.
\[\oint_c \vec F \cdot d \vec r = 0\]
Example
If \(\vec F = 2xyz \hat i + x^2 z \hat j + x^2 y \hat k\) then evaluate \(\int_c \vec F \cdot d \vec r\) between \(A(0, 0, 0)\) and \(B(2, 4,6)\).
- Along the curve \(c\) whose parametric equations are \(x = u\), \(y = u^2\) and \(z = 3u\)
- Along 3 straight
lines \(c_1: (0, 0, 0)\) to \((2, 0, 0)\), \(c_2: (2, 0, 0)\) to \((2, 4, 0)\) and \(c_3: (2, 4, 0)\) to \((2, 4, 6)\).
Solution
\[\vec F(x, y, z) = 2xyz \hat i + x^2 z \hat j + x^2 y \hat k\]
\[d \vec r = dx \hat i + dy \hat j + dz \hat k\]
\[\vec F(x, y, z) \cdot d \vec r = (2xyz \hat i + x^2z \hat j + x^2y \hat k) \cdot (dx \hat i + dy \hat j + dz \hat k)\]
\[=2xyz \, dx+x^{2}z \, dy+x^{2}y \,dz\]
\[\because x = u \implies dx = du\]
\[\because y = 2u \implies dy = 2du\]
\[\because z = 3u \implies dz = 3du\]
\[\therefore \vec F(u) \cdot d \vec r = 15u^4 du\]
Using any of the \(x(u), y(u), z(u)\), we can see that the bounds for \(u\) are \(0, 2\).
\[\therefore \int_c \vec F(u) \cdot d \vec r = \int_0^2 15u^4 du = \left[3u^5\right]_0^2 = 96\]
Now we will integrate along the straight lines.
\(c_1\)
\[c_1: (0, 0, 0) \text{ to } (2, 0, 0)\]
This shows us that
\[y = 0, z = 0, dy = 0\]
\[\int_{c_1} \vec F(x, y, z) \cdot d \vec r = \int_{c_1} (2xyz~dx+x^{2}zdy+x^{2}ydz)\]
\[= \int_{c_1}0 + 0 + 0 = 0\]
\(c_2\)
\[c_2: (2, 0, 0) \text{ to } (2, 4, 0)\]
This shows us that
\[x = 2, dx = 0, z = 0, dz = 0\]
\[\int_{c_2} \vec F(x, y, z) \cdot d \vec r = \int_{c_2} (2xyz~dx+x^{2}zdy+x^{2}ydz)\]
\[= \int_{c_2}0 + 0 + 0 = 0\]
\(c_3\)
\[c_3: (2, 4, 0) \text{ to } (2, 4, 6)\]
This shows us that
\[x = 2, dx = 0, y = 4, dy = 0\]
\[\int_{c_3} \vec F(x, y, z) \cdot d \vec r = 0 + 0 + \int_0^6 16 dz\]
\[= \left[16z\right]_0^6 = 96\]
\[\int_c \vec F \cdot d \vec r = \int_{c_1 + c_2 + c_3} \vec F \cdot d \vec r\]
\[= \int_{c_1} \vec F \cdot d \vec r + \int_{c_2} \vec F \cdot d \vec r + \int_{c_3} \vec F \cdot d \vec r\]
\[= 0 + 0 + 96 = 96\]
This shows that the vector field \(\vec F\) is conservative.
Also,
\[\nabla \times \vec F = \left|\begin{matrix}\hat i & \hat j & \hat k \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\2xyz & x^2 z & x^2y\end{matrix}\right|\]
\[= 0\]
We can apply 3 conditions to check if a vector field \(\vec F\) is conservative or not.
-
\[\oint \vec F \cdot d \vec r\]
-
\[\nabla \times \vec F = 0\]
-
\[\vec F = \nabla V\]
Divergence Theorem or Gauss' Theorem
For a closed surface \(S\), enclosing a region \(V\) in a vector field \(\vec F\),
\[\int_V \nabla \cdot \vec F dV = \int_S \vec F \cdot d \vec S\]
Example
Verify the divergence theorem for the vector field.
\[\vec F = x^2 \hat i + z \hat j + y \hat k\]
taken over the region bounded by the planes \(z = 0\), \(z = 2\), \(x = 0\), \(x = 1\), \(y = 0\) and \(y = 3\).
\[dV = dxdydz\]
We have to show that
\[\int_V \nabla \cdot \vec F dV = \int_S \vec F \cdot d \vec S\]
Left Hand Side
\[\nabla \cdot \vec F = \left(\frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z} \hat k\right) \cdot (x^2 \hat i + z \hat j + y \hat k)\]
\[=\frac{\partial}{\partial x}(x^{2})+\frac{\partial}{\partial y}(z)+\frac{\partial}{\partial z}(y)\]
\[=2x+0+0=2x\]
\[\therefore \int_V \vec F dV = \int_V 2x dV\]
\[= \iiint_V 2x dzdydx\]
\[= \int_0^1 \int_0^3 \int_0^2 2x dzdydx\]
\[= 6\]
Right Hand Side
\[\int_S \vec F \cdot d \vec S = \int_S \vec F \cdot \hat n dS\]
The \(S\) consists of \(6\) surfaces.
Base
\[S_1: z = 0, \hat n = - \hat k\]
\[\therefore \vec F = x^2 \hat i + y \hat k\]
\[\therefore dS_1 = dx dy\]
\[\int_{S_1} \vec F \cdot \hat n dS = \iint_{S_1} (x^2 \hat i + y \hat k) \cdot (- \hat k) dy dx\]
\[= \int_0^1\int_0^3 (-y) dy dx\]
\[= - \frac 9 2\]
Top
\[S_2: z = 2, \, \hat{n} = \hat{k}\]
\[\therefore \vec F = x^2 \hat i + z \hat j + y \hat k\]
\[\therefore \, dS_2 = dx \, dy\]
\[\int_{S_2} \vec F \cdot \hat{n} \, dS = \iint_{S_2} \left( x^2 \hat{i} + z \hat{j} + y \hat{k} \right) \cdot (\hat{k}) \, dy \, dx\]
\[= \int_0^1 \int_0^3 y \, dy \, dx = \frac{9}{2}\]
Right
\[S_3: y = 3, \hat n = \hat j \cdot dS_3 = dx \, dz \, \hat j\]
\[\therefore \vec F = x^2 \hat i + z \hat j + y \hat k\]
\[\int_{S_3} \vec F \cdot \hat{n} \, dS = \iint_{S_3} \left( x^2 \hat{i} + z \hat{j} + y \hat{k} \right) \cdot (\hat{j}) \, dz \, dx\]
\[= \int_0^1\int_0^2 z dz dx\]
\[= 2\]
Left
\[S_4: y = 0, \hat n = - \hat j\]
\[\therefore dS_4 = dx \, dz\]
\[\therefore \vec F = x^2 \hat i + z \hat j + y \hat k\]
\[\int_{S_4} \vec F \cdot \hat{n} \, dS = \iint_{S_4} \left( x^2 \hat{i} + z \hat{j} + y \hat{k} \right) \cdot (\hat{-j}) \, dz \, dx\]
\[= \int_0^1\int_0^2 (-z) dz dx\]
\[= -2\]
Front
\[S_5: x = 1, \hat n = \hat i\]
\[\therefore dS_5 = dy dz\]
\[\int_{S_5} \vec F \cdot \hat{n} \, dS = \iint_{S_5} \left(\hat{i} + z \hat{j} + y \hat{k} \right) \cdot (\hat{i}) \, dy \, dz\]
\[\int_0^2\int_0^3 1 \, dy \, dz\]
\[= 6\]
Back
\[x = 0, \hat n = - \hat i\]
\[\therefore dS_5 = dy \, dz\]
\[\int_{S_6} \vec F \cdot \hat{n} \, dS = \iint_{S_6} \left(z \hat{j} + y \hat{k} \right) \cdot (\hat{-i}) \, dy \, dz\]
\[= \iint_{S_6} 0 dy \, dz = 0\]
\[\therefore \int_S \vec F \cdot dS = - \frac 9 2 + \frac 9 2 + 2 - 2 + 6 + 6 + 0 = 6\]
Hence, verified.
Stoke's Theorem
If \(\vec F\) is a vector field existing over a surface \(S\) over a boundary \(c\) then
\[\int_S \nabla \times \vec F \cdot d \vec S = \oint_c \vec F \cdot d \vec r\]
Example
A hemisphere \(S\) is defined as \(x^2 + y^2 + z^2 = 4\) where \(z \ge 0\).
A vector field \(\vec F = 2y \hat i + x \hat j+ xz \hat k\) exists over the surface around its boundary \(c\).
Verify stokes theorem on it.
\[\int_S \nabla \times \vec F \cdot d \vec S = \oint_c \vec F \cdot d \vec r\]
\[S: x^2 + y^2 + z^2 - 4 = 0\]
\[c : x^2 + y^2 = 4\]
\[\vec F = 2y \hat i - x \hat j + xz \hat k\]
Right Hand Side
\[\oint_c \vec F \cdot d \vec r = \int_c (2y \hat i - x\hat j + xz \hat k) \cdot (\hat i dx + j \hat dy + \hat k dz)\]
\[= \int_c (2y dx - x dy + xz dz)\]
Converting coordinates into polar coordinates,
\[\because x = 2 \cos (\theta) \implies dx = - 2 \sin(\theta)\]
\[\because y = 2 \sin (\theta) \implies 2 \cos(\theta)\]
\[\because z = 0\]
The bounds for \(\theta\) will be
\[\theta = 0, 2\pi\]
Therefore, our integral becomes
\[= -4 \int_0^{2 \pi} (2 \sin^2(\theta) + \cos^2(\theta)) d\theta\]
\[= - 12 \pi\]
Left Hand Side
\[\int_S \nabla \times \vec F \cdot d \vec S = \int_S \nabla \times \vec F \cdot \hat n dS\]
\[\vec F = 2y \hat i - x \hat j + xz \hat k\]
\[\nabla \times \vec F = \left|\begin{matrix}\hat i & \hat j & \hat k \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\2y & -x & xz\end{matrix}\right|\]
\[= -z \hat j - 3 \hat k\]
\[\hat n = \frac{\nabla S}{\lvert \nabla S \rvert}\]
\[= \frac{2x \hat i + 2y \hat j + 2z \hat k}{\sqrt{4x^2 + 4y^2 = 4z^2}}\]
\[= \frac{x \hat i + y \hat j + z \hat k}{2}\]
\[\int_S \nabla \times \vec F \cdot \hat n dS = \int_S (-z \hat k - 3 \hat k) \cdot \left(\frac{x \hat i + y \hat j + z \hat k} 2\right) dS\]
\[= \frac 1 2 \int_S (- yz - 3z) dS\]
Converting the coordinates into spherical coordinate system.
\[x = 2 \sin (\phi) \cos (\theta)\]
\[y = 2 \sin (\phi) \sin (\theta)\]
\[z = 2 \cos (\theta)\]
\[dS = 4 \sin (\phi) d\phi \, d \theta\]
\[\therefore \int_S \nabla \times \vec F \cdot \hat n dS = - \int_0^{2 \pi}\int_0^{\frac \pi 2}(2\sin^{2}(\phi) \cos(\phi) \sin(\theta)+3\sin(\phi) \cos(\phi))d\phi d\theta\]
\[-12 \pi\]
Hence, verified.
References
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