## Integration over Vector Fields

I mentioned vector fields in a previous post in the context of differential equations and over the last week or so I have been looking at them in a bit more detail. Vector fields sound quite complicated but they can be very simple. A vector field can be presented visually as a vector attached to each point in space. The space may be the $x$-$y$ plane, three-dimensional space, it could be a region of the $x$-$y$ plane or even a manifold. A typical vector field in 2-dimensions might look as follows.

An image of a simple vector field drawn in SAGE Math

The arrows represent the vector that is attached to that particular point – the direction of the arrow gives the direction of the vector and the size of the arrow gives an idea of its relative magnitude. Graphical representations of vector fields can be a little misleading as it is tempting to think that only certain points have vectors attached to them – this is not the case; every point has a vector attached to it but if we were to try to show all of them the diagram would be too cluttered.

Vector fields are useful to model flows of liquids or gases; for example in weather prediction a vector field that changes over time could be used to model wind patterns. The vector attached to each point would tell you the direction and strength of the wind at that point and the vector field would evolve from one moment to the next. If you define a surface in a vector field then you can use integration to measure the flux across the surface – the physical interpretation of flux is as a measure of the amount of substance flowing across a surface. I came across this question in the book Advanced Engineering Mathematics Fifth Edition by Stroud and Booth and decided to give it a go.

Evaluate $\int_{S}\mathbf{F}.\mathrm{d}\mathbf{S}$ over the surface $S$ defined by $x^{2}+y^{2}+z^{2}=4$ for $z\geq0$ and bounded by $x=0$, $y=0$, $z=0$ and $\mathbf{F}=x\mathbf{i}+2z\mathbf{j}+y\mathbf{k}$.

The pictures below give an idea of what the vector field and surface both look like from a few different angles. This problem is asking to integrate this vector field over the surface to find the flux across the surface.

For this problem, since the surface that we are integrating over is part of a sphere, it is convenient to change to spherical polar co-ordinates given by$$x=r\mathrm{sin}\theta\mathrm{cos}\phi$$ $$y=r\mathrm{sin}\theta\mathrm{sin}\phi$$ $$z=r\mathrm{cos}\theta$$.The integration itself is quite straightforward although some of the integrands look a bit of a pain at first glance, but some techniques from A-Level Further Maths courses should clear things up. I used some reduction formulae to deal with some of the integrals that I ended up with which really simplified things (which is always good). You can download and view my full solution here – Integrating Vector Fields.

## Calculus of Residues

After blowing off the cobwebs after a couple of years I have been looking at some of the notes that I made some years back on some courses that I took at Warwick on Complex Analysis and Vector Analysis.

Integration has always been one of my favourite areas of mathematics. At A-Level I learned lots of different techniques for calculating some interesting integrals – but A-Level only just skims the surface when it comes to integration and it can be difficult for A-Level students (through no fault of their own) to appreciate the significance of integration. Integration by Parts, Integration by substitution and reduction formula are all great but there are still many integrals which require more advanced techniques to calculate. Contour integrals and the calculus of residues can often come to the rescue.

Contour integrals are a way of passing difficult integrals over a real-interval such as $$\int_{-\infty}^{\infty}\!{\dfrac{x^{2}}{1+x^{4}}\mathrm{d}x}$$ into the complex plane and taking advantage of the Cauchy integral theorem and the calculus of residues. I remember how it felt when I first learned the formula for integration by parts because it meant that I was able to find integrals that were previously impossible for me to calculate – even though I have done contour integrals before it has been very exciting for me to re-discover them. Looking through one of my books I came across this problem – show that for $a>1$

$$\int_{0}^{2\pi}\!\frac{\mathrm{sin}2\theta}{(a+\mathrm{cos}\theta)(a-\mathrm{sin}\theta)}\;\mathrm{d}\theta = -4\pi\left(1-\frac{2a\sqrt{a^{2}-1}}{2a^{2}-1}\right)$$

After spending a good deal of one of my afternoons wrestling with the algebra I managed to arrive at a solution which you can download here as a pdf. Here is a graph of the integrand in the case when $a=2$

As you can see from the diagram, the area bounded by the curve and the $x$-axis certainly exists but trying to calculate this integral using A-Level techniques is going to be incredibly difficult if not impossible (if anyone can do it then I would love to see the solution). Unfortunately there are and always will be integrals that cannot be calculated analytically – this is just the way it is and there is no getting around it but contour integrals certainly allows you to calculate a huge range of integrals that previously would have been seemingly impossible.