Chapter 1: Integral Calculus

Section 3.1: Line, Surface, Volume Integrals

  1. Line Integral - At each point on the path, take a dot product between \(\boldsymbol{v}\) and infinitesimal displacement vector along the path \(\partial{\boldsymbol{l}}\) => \( \int_a^b v \cdot \partial{l} \).
    For closed loop (starting and final coordinates are equal b=a) use \(\oint_a^b v \cdot \partial{l} \)
    Ordinarilly, line(or path) integral depends on path taken. If line integral's values is independent of the path , determined entirely by the end points -> vector function is conservative.
  2. Surface Integral - \( \int_S v \cdot \partial{a} \), \(\boldsymbol{v}\) is vector function, \(\partial{a} = \partial{x}\partial{y}\) is an infinitesimal patch of area with direction perpendicular to the surface.
    Outward \(\partial{a}\) is positive, but is ambiguous for open surfaces.
    If \(\boldsymbol{v}\) describes the flow of a fluid (mass per unit area per time), then \( \int_S v \cdot \partial{a} \) describes the total mass per unit area per time passing through surface \(S\). Alternative name -> flux.
    Value depends on particular surface chosen , BUT there exist a special class of functions for which it is independent of the surface, only by boundary line.
  3. Volume Integral - \( \int_V T \cdot \partial{\tau} \), where T - scalar or vector function, \(\partial{\tau} = \partial{x}\partial{y}\partial{z}\) is infinitesimal volume element.
    If T is the density of a substance, then volume integral is the total mass.
Relevant problems:
1.29: Calculate the line integral of the function \(\boldsymbol{v} = x^2 \boldsymbol{\hat{x}} + 2yz \boldsymbol{\hat{y}} + y^2 \boldsymbol{\hat{z}}\) from the origin to the point (1,1,1) by three different routes:

Write down definition of a line integral. Think about each piece and how you could simplify each dot product (ex why are you being asked to use paths along the x, y and z axis? how does that simplify your line integral?)

1.31: Find the volume integral of the function \(T=z^2\) over a tetrahedron with corners (0,0,0) (0,1,0) (1,0,0) (0,0,1)

Draw the tetrahedron. Hard part is to find limits of integration. However, to find limits of integration you need to know how to find equation of a plane. I suggest reading over Paul's Notes: Triple Integrals and Equations of Planes and and then attempting the problem.

  1. To find the normal to the plane of tetrahedron we need to define vectors laying on the plane and take their cross product: \(\boldsymbol{A} = \hat{x} - \hat{z}\) and \(\boldsymbol{B} = \hat{y} - \hat{z}\)
  2. \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 0 & 1 & -1 \\ \end{vmatrix} = \hat{i} + \hat{j} + \hat{z}\)
  3. Equation of a plane: \(xslope(x - x_1) + yslope(y - y_1) + zslope(z-z_1) = 0\)
  4. Substitute point (0,0,1) => \((x - 0) + (y - 0) + (z-1) = 0 \rightarrow z=1-x-y\)
  5. Easiest to integrate with respect to z last: \(\int_0^1 z^2 \int\int_{u_1(y,z)}^{u_2(y,z)}\partial{x}\partial{y}\partial{z} = \int_0^1 z^2 \int\int_{0}^{x=1-y-z}\partial{x}\partial{y}\partial{z} = \int_0^1 z^2 \int_{u_1(z)}^{u_2(z)} (1-y-z)\partial{y}\partial{z}\)
  6. Most intuitive limits: \(0 < z < 1, 0 < y < 1 - \underbrace{x}_{x = 0} - z\)
  7. \(\int_0^1 z^2 \int_{0}^{1 - z} (1-y-z)\partial{y}\partial{z} =\int_0^1 z^2 [(1 - z)^2 - \frac{(1 - z)^2}{2}]\partial{z} = \int_0^1 z^2 [\frac{(1 - z)^2}{2}]\partial{z} = \int_0^1 z^2 [\frac{1}{2} - z + \frac{z^2}{2}]\partial{z} = \frac{1}{6} - \frac{1}{4} + \frac{1}{10} = \frac{1}{60}\)
Alternative way to integrate:

Section 3.2: Fundamental Theorem Of Calculus For Gradients

Fundamental Theorem of Calculus: Integral of a derivative over an interval is given by the value of the function at the end points. Read more at Paul's notes.
FTC for gradients: If T(x,y,z) is a scalar function, starting at point a, move \(\partial{l}\) function T will change by \(dT = (\nabla{T})\partial{l}\)
Total change in T going from a to b along a selected path: \(\require{enclose} \enclose{box}[mathcolor="black"]{\int_a^b (\nabla{T})\partial{l} = T(b) - T(a)}\)
Special properties:

  1. \(\int_a^b (\nabla{T})\partial{l}\) is independent of the path taken from a to b
  2. \(\oint (\nabla{T})\partial{l} = 0 \) since begininng and end points are identical. T(b) - T(b) = 0

Section 3.3: Fundamental Theorem Of Calculus For Divergences

Fundamental Theorem for Divergences: \(\enclose{box}[mathcolor="black"]{\int_V (\nabla \cdot \boldsymbol{v})\partial{\tau} = \oint_S \boldsymbol{v} \cdot \partial{a}}\) (Gauss's, Green's, divergence theorem)
Geometric interpretation: If \(v\) represents the flow of an incompresible fluid then the flux of \(\boldsymbol{v}\) (right side) is the total amount of fluid passing out through the surface per unit time.
\(\int_V \text{faucets within the volume} = \oint \text{flow out through the surface area}\)

Section 3.4: Fundamental Theorem For Curls (Stokes Theorem)

Integral of a derivative over a region (patch of surface) is equal to the value of the function at the boundary. \(\enclose{box}[mathcolor="black"]{\int_S (\nabla \times \boldsymbol{v}) \partial{a} = \oint_P \boldsymbol{v} \cdot \partial{l}}\)

  1. \(\int (\nabla \times \boldsymbol{v}) \partial{a}\) depends only on the boundary line, not on the particular surface used.
  2. \(\oint (\nabla \times \boldsymbol{v}) \partial{a} = 0\) for any closed surface, since the boundary line, like baloon mouth, shrinks down to a point (linear integral evaluates to 0).
Relevant problems:
1.35: Check Corollary 1 by using the same function and boundary line as in Ex. 1.11, but integrating over the five faces of the cube in Fig. 1.35. The back of the cube is open.

Curl of the function was already calculated in Ex 1. Find out unit area vectors for each of the surfaces and calculate its dot product with the curl.

Section 3.5: Integration by Parts

Exploit product rules: \(\nabla \cdot (f\boldsymbol{A})) = f(\nabla \cdot \boldsymbol{A}) + \boldsymbol{A} \cdot \nabla{f} \rightarrow \underbrace{\int_V \nabla \cdot (f\boldsymbol{A}) \partial{\tau} = \int_S f\boldsymbol{A} \cdot \partial{a}}_{\text{divergence theorem}} = \int_V f(\nabla \cdot \boldsymbol{A}) \partial{\tau} + \int_V \boldsymbol{A} \cdot \nabla{f} \partial{\tau}\)
Thus: \(\int_V f(\nabla \cdot \boldsymbol{A}) \partial{\tau} = -\int_V \boldsymbol{A} \cdot \nabla{f} \partial{\tau} + \underbrace{\int_S f\boldsymbol{A} \cdot \partial{a}}_{\text{boundary term}}\)

Relevant problems:
1.36: Show that:
  1. \(\int_S f(\nabla \times \boldsymbol{A}) \partial{\tau} = \int_S [\boldsymbol{A} \times (\nabla{f})] \cdot \partial{a} + \oint_P f\boldsymbol{A} \cdot \partial{l} \)
  2. \(\int_V \boldsymbol{B} \cdot (\nabla \times \boldsymbol{A}) \partial{\tau} = \int_V \boldsymbol{A} \cdot (\nabla \times \boldsymbol{B}) \partial{\tau} + \oint_S (\boldsymbol{A} \times \boldsymbol{B}) \cdot \partial{a}\)

Look at vector calculus product rules in Griffith. Rules v and iv are particularly helpful.

  1. Rule V: \(\nabla \times (f\boldsymbol{A}) = f(\nabla \times \boldsymbol{A}) - \boldsymbol{A} \times (\nabla{f})\)
    \(\int_S f(\nabla \times \boldsymbol{A}) \cdot \partial{a} = \int_S \nabla \times (f\boldsymbol{A}) \cdot \partial{a} + \int_S \boldsymbol{A} \times (\nabla{f}) \cdot \partial{a} \rightarrow \int_S f(\nabla \times \boldsymbol{A}) \cdot \partial{a} = \underbrace{\oint_P f\boldsymbol{A} \cdot \partial{l}}_{\text{Stokes' Theorem}} + \int_S \boldsymbol{A} \times (\nabla{f}) \cdot \partial{a}\)
  2. Rule IV: \(\nabla \cdot (\boldsymbol{A} \times \boldsymbol{B})= \boldsymbol{B} \cdot (\nabla \times \boldsymbol{A}) - \boldsymbol{A} \cdot (\nabla \times \boldsymbol{B})\)
    \(\int_V \nabla \cdot (\boldsymbol{A} \times \boldsymbol{B}) \partial{\tau} = \int_V \boldsymbol{B} \cdot (\nabla \times \boldsymbol{A}) \partial{\tau} - \int_V \boldsymbol{A} \cdot (\nabla \times \boldsymbol{B}) \partial{\tau} \rightarrow \int_V \boldsymbol{B} \cdot (\nabla \times \boldsymbol{A}) \partial{\tau} = \int_V \nabla \cdot (\boldsymbol{A} \times \boldsymbol{B}) \partial{\tau} + \int_V \boldsymbol{A} \cdot (\nabla \times \boldsymbol{B}) \partial{\tau} \)
    \(\int_V \boldsymbol{B} \cdot (\nabla \times \boldsymbol{A}) \partial{\tau} = \underbrace{\oint_S \boldsymbol{A} \times \boldsymbol{B} \cdot \partial{a}}_{\text{Divergence Theorem}} + \int_V \boldsymbol{A} \cdot (\nabla \times \boldsymbol{B}) \partial{\tau}\)