1) Introduction: Exploring Vector Fields š
Vector calculus extends integration to vector fields, introducing line integrals along curves and surface integrals over surfaces, unified by theorems like Greenās Theorem and Stokesā Theorem. These concepts are foundational in physics (e.g., electromagnetism, fluid flow) and engineering, connecting 2D and 3D analyses. This topic provides a comprehensive guide to mastering these techniques.
Weāll cover:
- Line Integrals: Work and circulation along paths.
- Surface Integrals: Flux and area over surfaces.
- Greenās Theorem: Linking line and double integrals.
- Stokesā Theorem: Extending to 3D surfaces.
Quick Recap: Multiple integrals handled scalar functions; now we integrate vector quantities.
2) Line Integrals š
A line integral computes work or circulation along a curve \( C \) defined by a parameterization.
Definition 30.1: Line Integral
For a vector field \( \mathbf{F} = (P, Q, R) \) and curve \( C \) parameterized by \( \mathbf{r}(t) = (x(t), y(t), z(t)) \), \( t \in [a, b] \), the line integral is \( \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt \). For scalar functions, \( \int_C f \, ds \) uses arc length \( ds = |\mathbf{r}'(t)| \, dt \).
Example 1: Scalar Line Integral
Evaluate \( \int_C y \, ds \) where \( C \) is the parabola \( y = x^2 \), \( 0 \leq x \leq 1 \).
- Parameterize: \( x = t \), \( y = t^2 \), \( dx = dt \), \( dy = 2t \, dt \), \( ds = \sqrt{1 + (2t)^2} \, dt \).
- \( \int_0^1 t^2 \sqrt{1 + 4t^2} \, dt \).
- Substitute \( u = 1 + 4t^2 \), \( du = 8t \, dt \), adjust limits, compute.
- Answer: \( \frac{1}{12}(5\sqrt{5} - 1) \).
Answer: \( \frac{1}{12}(5\sqrt{5} - 1) \).
Example 2: Vector Line Integral
Compute \( \int_C (y, -x) \cdot d\mathbf{r} \) where \( C \) is \( \mathbf{r}(t) = (\cos t, \sin t) \), \( 0 \leq t \leq \pi \).
- \( \mathbf{r}'(t) = (-\sin t, \cos t) \).
- \( \mathbf{F} = (\sin t, -\cos t) \), \( \mathbf{F} \cdot \mathbf{r}' = -\sin t \cdot (-\sin t) + (-\cos t) \cdot \cos t = \sin^2 t - \cos^2 t \).
- \( \int_0^{\pi} (\sin^2 t - \cos^2 t) \, dt = \int_0^{\pi} -\cos 2t \, dt = [-\frac{\sin 2t}{2}]_0^{\pi} = 0 \).
Answer: \( 0 \).
Example 3: Work Calculation
Find work done by \( \mathbf{F} = (x, y) \) along \( y = x^2 \) from \( (0, 0) \) to \( (1, 1) \).
- \( \mathbf{r}(t) = (t, t^2) \), \( \mathbf{r}' = (1, 2t) \).
- \( \int_0^1 (t, t^2) \cdot (1, 2t) \, dt = \int_0^1 (t + 2t^3) \, dt = \left[\frac{t^2}{2} + \frac{t^4}{2}\right]_0^1 = \frac{1}{2} + \frac{1}{2} = 1 \).
Answer: \( 1 \).
3) Surface Integrals š
Surface integrals measure flux or surface area over a surface \( S \).
Definition 30.2: Surface Integral
For \( \mathbf{F} = (P, Q, R) \) and surface \( S \) parameterized by \( \mathbf{r}(u, v) \), the flux is \( \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, dA \). For area, \( \iint_S dS = \iint_D \|\mathbf{r}_u \times \mathbf{r}_v\| \, dA \).
Example 4: Surface Area
Find the area of \( z = x^2 + y^2 \) above \( x^2 + y^2 \leq 1 \).
- \( \mathbf{r}(u, v) = (u \cos v, u \sin v, u^2) \), \( u \in [0, 1] \), \( v \in [0, 2\pi] \).
- \( \mathbf{r}_u = (\cos v, \sin v, 2u) \), \( \mathbf{r}_v = (-u \sin v, u \cos v, 0) \).
- \( \mathbf{r}_u \times \mathbf{r}_v = (2u^2 \cos v, 2u^2 \sin v, -u) \), \( \|\cdot\| = u \sqrt{1 + 4u^2} \).
- \( \int_0^{2\pi} \int_0^1 u \sqrt{1 + 4u^2} \, du \, dv \).
Answer: \( \frac{\pi}{6}(17\sqrt{17} - 1) \).
Example 5: Flux Through a Surface
Compute \( \iint_S (0, 0, z) \cdot d\mathbf{S} \) over \( z = 1 - x^2 - y^2 \), \( z \geq 0 \).
- \( \mathbf{r} = (u \cos v, u \sin v, 1 - u^2) \), normal \( \mathbf{n} = (-2u \cos v, -2u \sin v, 1) \).
- Flux = \( \int_0^{2\pi} \int_0^1 (1 - u^2) \cdot 1 \, du \, dv = 2\pi \int_0^1 (1 - u^2) \, du = 2\pi \left[u - \frac{u^3}{3}\right]_0^1 = 2\pi \cdot \frac{2}{3} \).
Answer: \( \frac{4\pi}{3} \).
Example 6: Parametrized Surface
Evaluate \( \iint_S \mathbf{F} \cdot d\mathbf{S} \) where \( \mathbf{F} = (y, x, z) \), \( S: z = \sqrt{x^2 + y^2} \), \( x^2 + y^2 \leq 1 \).
- Parameterize, compute normal, integrate.
Answer: \( \frac{\pi}{2} \).
4) Greenās Theorem š
Greenās Theorem connects line integrals around a closed curve to double integrals over the enclosed region.
Definition 30.3: Greenās Theorem
\( \oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \), where \( C \) is a simple closed curve and \( R \) is the region it encloses.
Example 7: Applying Greenās Theorem
Evaluate \( \oint_C (x^2 - y^2) \, dx + (x + y) \, dy \) where \( C \) is \( x^2 + y^2 = 1 \).
- \( P = x^2 - y^2 \), \( Q = x + y \).
- \( \frac{\partial Q}{\partial x} = 1 \), \( \frac{\partial P}{\partial y} = -2y \).
- \( \iint_R (1 + 2y) \, dA \), use polar: \( y = r \sin \theta \).
Answer: \( 0 \) (symmetric).
Example 8: Verification
Compute \( \oint_C (-y \, dx + x \, dy) \) for \( x^2 + y^2 = 4 \).
- \( \frac{\partial}{\partial x}(x) = 1 \), \( \frac{\partial}{\partial y}(-y) = -1 \).
- \( \iint_R (1 - (-1)) \, dA = 2 \cdot \pi \cdot 2^2 = 8\pi \).
Answer: \( 8\pi \).
5) Stokesā Theorem š
Stokesā Theorem generalizes Greenās Theorem to surfaces in 3D.
Definition 30.4: Stokesā Theorem
\( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \), where \( C \) is the boundary of surface \( S \).
Example 9: Stokesā Application
Evaluate \( \oint_C (y, 0, 0) \cdot d\mathbf{r} \) where \( C \) is \( x^2 + y^2 = 1 \), \( z = 0 \), with \( S \) as \( z = 1 - x^2 - y^2 \).
- \( \nabla \times \mathbf{F} = (0, 0, -1) \).
- \( \iint_S (-1) \, dS = -\pi \).
Answer: \( -\pi \).
Example 10: 3D Surface
Compute \( \oint_C (z, x, y) \cdot d\mathbf{r} \) for \( C \) bounding \( z = x^2 + y^2 \), \( z \leq 1 \).
- \( \nabla \times \mathbf{F} = (0, 0, 2) \), integrate over \( S \).
Answer: \( 2\pi \).
6) Practice Questions šÆ
Fundamental Practice Questions š±
Instructions: Evaluate the line or surface integrals. š
\( \int_C x \, ds \) where \( C \) is \( y = x^3 \), \( 0 \leq x \leq 1 \)
\( \int_C (y, x) \cdot d\mathbf{r} \) where \( C \) is \( x = t^2 \), \( y = t \), \( 0 \leq t \leq 1 \)
\( \iint_S z \, dS \) where \( S \) is \( z = x + y \), \( 0 \leq x, y \leq 1 \)
\( \oint_C (x^2, y^2) \cdot d\mathbf{r} \) where \( C \) is \( x^2 + y^2 = 1 \)
\( \iint_S (0, 0, 1) \cdot d\mathbf{S} \) where \( S \) is \( z = 4 - x^2 - y^2 \), \( z \geq 0 \)
\( \int_C (x, 2y) \cdot d\mathbf{r} \) where \( C \) is \( y = \sin x \), \( 0 \leq x \leq \pi \)
\( \iint_S \mathbf{F} \cdot d\mathbf{S} \) where \( \mathbf{F} = (x, y, z) \), \( S: z = x^2 + y^2 \), \( z \leq 1 \)
\( \oint_C (y, -x) \cdot d\mathbf{r} \) where \( C \) is \( x^2 + y^2 = 9 \)
\( \iint_S (x, y, 0) \cdot d\mathbf{S} \) where \( S \) is \( z = \sqrt{1 - x^2 - y^2} \)
\( \int_C (xy, x^2) \cdot d\mathbf{r} \) where \( C \) is \( y = x^2 \) from \( (0, 0) \) to \( (2, 4) \)
\( \iint_S (0, 0, x) \cdot d\mathbf{S} \) where \( S \) is \( x + y + z = 1 \), \( x, y, z \geq 0 \)
Challenging Practice Questions š
Instructions: Solve these advanced vector calculus problems. š§
Evaluate \( \oint_C (x^3, y^3) \cdot d\mathbf{r} \) where \( C \) is the boundary of \( x^2 + y^2 \leq 4 \), using Greenās Theorem.
Compute \( \iint_S (z^2, xz, y) \cdot d\mathbf{S} \) where \( S \) is \( z = 1 - x^2 - y^2 \), \( z \geq 0 \), using Stokesā Theorem.
Find the flux of \( \mathbf{F} = (x^2, y^2, z^2) \) through \( x^2 + y^2 + z^2 = 4 \), \( z \geq 0 \).
Determine \( \oint_C (e^x, \sin y) \cdot d\mathbf{r} \) for \( C \) bounding \( y = x^2 \), \( 0 \leq x \leq 1 \).
Evaluate \( \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \) where \( \mathbf{F} = (yz, xz, xy) \), \( S \) is \( z = 1 - x^2 - y^2 \).
7) Summary & Cheat Sheet š
7.1) Line Integral
\( \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F} \cdot \mathbf{r}' \, dt \).
7.2) Surface Integral
\( \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, dA \).
7.3) Greenās Theorem
\( \oint_C (P \, dx + Q \, dy) = \iint_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA \).
7.4) Stokesā Theorem
\( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \).
Youāve mastered vector calculus! Next, weāll explore complex analysis. š