I have a vector field (velocity) and I want to integrate around a rectangle. If I had a function for the field it would be easy and I know how to do it. However, I only have the coordinates. Is there an easy way of doing this?
I thought about how I'm going to solve it, started writing the steps for the solution: parametrise each line, find the derivative of the parametrisation. However, I got stuck because in the integral, the field has to be evaluated at the parametric function.
$$\begin{align}
\int_C \vec F \cdot d\vec r &= \int_C \vec F \cdot \vec T \: ds \\
&= \int_a^b \vec F\left(\vec r(t)\right) \cdot \frac{\vec r'(t)}{\left\lVert \vec r'(t) \right\rVert} \left\lVert \vec r'(t) \right\rVert \: dt \\
&= \int_a^b \vec F\left(\vec r(t)\right) \cdot \vec r'(t) \: dt
\end{align}$$
The first part of the last line is where I got stuck, t would be involved and I don't know how to do it numerically.
***** Side note, I'm trying to calculate circulation which can also be found by calculating the surface integral of the curl of velocity (vorticity), but I guess that would be even harder *numerically***
I use MatLab but anything that would get me started would do
Thanks
Edit: The above equation was copied from a website, let me use my own notation:
I have a vector field $\vec V = \vec V(x,y)$ which has an x component $u$ and a y component $v$.
$\Rightarrow \vec V(x,y) = u(x,y) \; \vec i + v(x,y) \; \vec j $
I want to integrate along C which is a rectangle so I will have to divide the integration into 4 parts (4 smooth curves).
The vertices of the rectangle are $(0.0274,0.0386)\: (0.0926,0.0386) \: (0.0926,0.1184) \: (0.0274,0.1184)$
So the four curves are:
$$\begin{align}
C_1: \: \: r_1(t) = t \; \vec i + 0.0386 \; \vec j \hspace{2cm} for\hspace{1cm} 0.0274 \le t \ge 0.0926 \\
C_2: \: \: r_2(t) = 0.0926 \; \vec i + t \; \vec j \hspace{2cm} for\hspace{1cm} 0.0386 \le t \ge 0.1184 \\
-C_3: \: \: r_3(t) = t \; \vec i + 0.1184 \; \vec j \hspace{2cm} for\hspace{1cm} 0.0274 \le t \ge 0.0926 \\
-C_4: \: \: r_4(t) = 0.0274 \; \vec i + t \; \vec j \hspace{2cm} for\hspace{1cm} 0.0386 \le t \ge 0.1184
\end{align}$$
and the derivatives:
$$\begin{align}
r_1'(t) &= 1 \; \vec i + 0 \; \vec j \\
r_2'(t) &= 0 \; \vec i + 1 \; \vec j \\
r_3'(t) &= 1 \; \vec i + \; \vec j \\
r_4'(t) &= 0 \; \vec i + 1 \; \vec j
\end{align}$$
then:
$$\begin{align}
\Gamma = \int_C \vec V(x,y) \cdot d\vec r &= \\
&= \int_{C_1} \vec V(x,y) \cdot d\vec r + \int_{C_2} \vec V(x,y) \cdot d\vec r – \int_{-C_3} \vec V(x,y) \cdot d\vec r – \int_{-C_4} \vec V(x,y) \cdot d\vec r \\
&= \int^{b_1}_{a_1} \vec V(\vec r_1(t)) \cdot r_1'(t) dt + \int^{b_2}_{a_2} \vec V(\vec r_2(t)) \cdot r_2'(t) dt \\ &- \int^{b_3}_{a_3} \vec V(\vec r_3(t)) \cdot r_3'(t) dt -\int^{b_4}_{a_4} \vec V(\vec r_4(t)) \cdot r_4'(t) dt
\end{align}$$
This is where I got to, I know that $\vec V(\vec r(t)) = \vec V(x(t),y(t))$, here I cannot think of a way of integrating numerically (mainly the $t$ term in $r(t)$ confuses me). $x$ and $y$ are 2-D grids (a 41 by 35 matrix) where the columns in x are equal and the rows in y are equal. x-component $u(x,y)$ is also a 41 by 35 matrix and so is the y-component $v(x,y)$
Best Answer
Your work looks pretty good, but you can actually take it a step further since you know $r_i'(t)$. When you substitute in this information, each integral depends only on one component of $\vec V$, but not both. For instance $$\int^{b_1}_{a_1} \vec V(\vec r_1(t)) \cdot r_1'(t) \; dt =\int^{b_1}_{a_1} u(\vec r_1(t)) \; dt$$
The next task is to write a routine to implement the function $\vec V$, that is, you want a function from $\mathbb{R}^2\rightarrow\mathbb{R}^2$. Use something like
interp2(though this may not be the most efficient). Once you have this accomplished, simply evaluate the now 1-d integrals using thequadfunction, for instance.