Vector Calculus

I. Double Integration

Take a function f(x,y) over a region R in the x,y plane.

**Diagram 1

At a point  P, let us take an area ‘piece’ of of area with area \delta A_{p}. We then weigh each area piece by the value of the function at that point, f(x_{p},y_{p}), and take the sum over the region R.

    $$\sum_{p}f(x_{p},y_{p})\delta A_{p}$$

We recognize this to be a Reimann sum. Then consider the limit as \delta A_{p} \rightarrow 0. This corresponds to taking smaller and smaller pieces of the region R. This gives us a double integral, thus integrating over both x and y.

    $$I= \iint f(x,y) dA = \iint f(x,y) dxdy$$

When performing double integrals the weighted sum of sum function if taken over a region. The bounds of this region are used as the bounds of the integral, so that the weighted sum is taken over the relevant region.

II. Change of Variables

Change of variables are often used in integration.

Changes of variables are often used to simplify calculations or to change co-ordinate systems in order to take advantages of symmetries within systems. Each change of variable can be broken down into 3 main steps:

  • Express function in terms of new variable(s)
  • Change differential element(s)
  • Change limits to be in terms of new variable(s)

To illustrate this, let us consider once again a function of two variables f(x,y). We’ll consider a change of variables in the context of a double integral taking the weighted sum of f over a region.

    $$\iint f(x,y) dxdy$$

Let’s consider the following change of variables.

    $$x \rightarrow x(u,v)$$

    $$y \rightarrow y(u,v)$$

We now express f as f(x(u,v), y(u,v)), which is the first step of changing variables. The next step is to change our differential elements dx, dy.

To do so we consider the position vector of a point k in the plane (x_{k},y_{k}).

**Diagram 2

    $$\underline{r}= x\underline{\hat{i}}+y\underline{\hat{j}}$$

    $$\underline{dr}= dx\underline{\hat{i}}+dy\underline{\hat{j}}$$

The vector \underline{dr} represents a small change along constant x and a small change along a constant y.

**Diagram 3

Here the vector \underline{KL} represents a small change u+du, and the vector \underline{KN} represents a small change v+dv. The area spanned by the vectors \underline{KL} and \underline{KN} corresponds to the new area element. Therefore we want to find this area.

Using the total derivative we can express the differentials dx and dy in terms of du and dv. This allows us to find the vectors \underline{KL} and \underline{KN} as they represent changes along \underline{dr} at constant v and u respectively.

Let’s find \underline{KL}, as \underline{KN} follows similarly. Along \underline{KL} v is constant, therefore the total derivatives dx and dy simplify.

    $$dx=\frac{\partial x}{\partial u}du + \cancelto{0}{\frac{\partial x}{\partial v}dv}$$

    $$dy=\frac{\partial y}{\partial u}du + \cancelto{0}{\frac{\partial y}{\partial v}dv}$$

Thus \underline{KL} is the vector \frac{\partial x}{\partial u}du \ \underline{\hat{i}} +\frac{\partial y}{\partial u}du \ \underline{\hat{j}} , and similarly \underline{KL} is the vector \frac{\partial x}{\partial v}dv \ \underline{\hat{i}} +\frac{\partial y}{\partial v}dv \ \underline{\hat{j}}. The area spanned by these vectors is the area we are looking for. Recall that the modulus of the cross product of two vectors is equal to the area spanned, so the area can be found.

    $$dxdy= A= \left | \underline{KL} \times \underline{KN} \right |  $$

    $$=\begin{vmatrix} \underline{\hat{i}} & \underline{\hat{j}} & \underline{\hat{k}}\\ \frac{\partial x}{\partial u}du& \frac{\partial y}{\partial u}du& 0\\ \frac{\partial x}{\partial v}dv& \frac{\partial y}{\partial v}dv& 0 \end{vmatrix}= \begin{vmatrix} \frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v}& \frac{\partial y}{\partial v} \end{vmatrix}dudv$$

This gives us the area element dxdy in terms of the new variables.

    $$dxdy=\begin{vmatrix} \frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}\\ \frac{\partial x}{\partial v}& \frac{\partial y}{\partial v} \end{vmatrix}dudv = \left | J \right | dudv$$

Here \left |J \right| is called the Jacobian.

We can now express the double integral in terms of the change of variables, as the function has been expressed in terms of the new variables, and now the area element has also been expressed in terms of the new variables. In practice, when an area is given over which to take the integral a change in limits is also performed.

    $$\iint f(x,y)dxdy= \iint f(x(u,v), y(u,v))\left |J \right|  dudv$$

III. Triple Integration

Just as we interpreted the double integral as the weighted sum over an area, the triple integral is a weighted sum over a volume. A similar Reimann sum can be derived, and in the infinitesimal limit this tends to the triple integral.

    $$\iiint f(x,y,z)dV $$


The change of variables here is derived in a similar way to the change of variables for a double integral. We consider this time a change of 3 variables. This means that we find a relation between small changes in the initial variables (usually denoted dx,dy,dz in 3 dimensions) and the vectors along small changes in the new variables. Once this relation has been found the new volume element can be found by the triple scalar product between the aforementioned vectors in a right hand set. This leads to the Jacobian in three dimensions.

    $$\begin{vmatrix} J \end{vmatrix}=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u}\\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v}\\ \frac{\partial x}{\partial w} & \frac{\partial y}{\partial w} & \frac{\partial z}{\partial w} \end{vmatrix}$$

This means that the fully transformed integral is in the following form.

    $$\iiint f(x(u,v,w), y(u,v,w), z(u,v,w)) \begin{vmatrix}J\end{vmatrix} dudvdw$$

This Jacobian can be used to find the volume element in both Spherical and Cylindrical polar co-ordinates. Using the relation between cartesian co-ordinates and the various \mathbb{R}^{3} polar co-ordinates, the relevant partial derivatives.