Integral on a curve $\int_C (x^2+y^2)ds$

Question

Calculate $\int_C(x^2+y^2)ds$ where $C$ is the segment $[AB]$, $A(a,a)$, $B(b,b), b > 0$. How could I parametrize this curve and can you also explain your thinking process while doing so?

UPDATE:

I tried using the parametrization $x = t, y =t$, $t\in [a,b]\implies \int_C(x^2+y^2)ds=\int_a^b2t^2dt=\frac 23 (b^3-a^3).$ But I'm not sure if it's correct what i did

Answer 1

It is not correct. We have the parametrisation $c(t)=(t,t)$, hence $c'(t)=(1,1)$, thus $||c'(t)||=\sqrt{2}.$

Therefore $\int_C(x^2+y^2)ds=\int_a^b2 \sqrt{2}t^2dt.$

Answer 2

So there are two scenarios that come to mind. The first is where you are given both $x$ and $y$ as functions of a common variable, for example: $x(t)$ and $y(t)$. The second scenario is where you are given a more direct relation, for example: $y(x)$.

Preliminaries:

So, before I start with the methods of solution, I will explain how $ds$ relates to $dx$ and $dy$.

If you move along any arbitrary, yet continuous, curve C, between two points that are relatively close, you can approximate the distance travelled as $\Delta s$ which can be calculated as follows:

$$\Delta s^2 = \Delta x^2 + \Delta y^2$$

Now, this is not the actual distance, as you might have gone around a curve which could be longer than this shortest distance between the departure and arrival points. Now, as you move the arrival point closer to the departure point, this curved path starts to look more and more like the shortest path between the two points, that is the shortest path $\Delta s$ starts to approximate this distance. If we let this continue, the distance will become infinitesimally small and we usually denote this distance as $ds$, which is also called a differential, which can then be calculated from the two other differentials $dx$ and $dy$ as follows:

$$ds^2 = dx^2 + dy^2$$

Now, along this curve, at each point on the curve, there is some value or parameter in which we might be interested. For instance, we might be interested in the fuel left in our airplane, which would vary as we fly from point to point. The rate, per kilometer for example, at which the airplane consumes fuel might be known and can be represented as a function of the two variables $x$ and $y$, or more commonly as a function of the intermediate variable $t$ which we will imagine to be time in this scenario. Lets denote this as follows:

$$f(x(t), y(t))$$

Next, we might want to ask a question, such as "how much fuel have we used to get from some point $A = (x_a, y_a)$ to another point $B = (x_b, y_b)$". The distance, along the curve is measured in kilometers. So, if we multiply the rate of fuel consumption, $f(x,y)$ with the total distance, we will get the total fuel used during the flight.

But, this could be tricky, since the fuel consumption rate could change as the time goes by, so instead we will measure the fuel consumption, fly a short distance, and multiply the measured consumption with the short distance. We can then repeat this process over and over again and simply add all of these intermediate results to get a good approximation of the total fuel consumed. If we shorten the distance over which we do this, this distance again approaches zero, and can be represented by the differential $ds$. We can represent the solution to this question as follows:

$$fuel_{total} = \int_{A}^{B}f(x,y)ds$$

Parametrization:

So, we start with the equation involving the differentials:

$$ds^2 = dx^2 + dy^2$$

We are actually only interested in $ds$, so we can rewrite it in all the following ways:

\begin{align*} ds &= (dx^2 + dy^2)^{\frac{1}{2}} \\ ds &= \left(1 + \left(\frac{dy}{dx}\right)^2\right)^{\frac{1}{2}} dx\\ ds &= \left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right)^{\frac{1}{2}} dt\\ \end{align*}

Solution:

In your situation we see that $x = y$ is an intrinsic relation. We can choose $x$ as the intermediate variable. We then also have to rewrite the bounds on the integral as the corresponding $x$ values, $a$ and $b$. Since we have chosen $x$ as the intermediate variable, we can use the second of the three expressions of $ds$ above to solve this problem. We have also been given the fuel consumption rate as $f(x,y) = x^2 + y^2 = x^2 + x^2 = 2x^2$ So we can rewrite and reformulate the question as follows:

\begin{align*} total_{fuel} &= 2\int_{a}^{b} x^2 ds \\ total_{fuel} &= 2\int_{a}^{b} x^2 \left(1 + \left(\frac{dy}{dx}\right)^2\right)^{\frac{1}{2}} dx \\ \end{align*}

Now, we can easily find $\frac{dy}{dx}$ in this case. It is simply $1$. So we can substitute this intermediate result into the question and simplify as follows:

\begin{align*} total_{fuel} &= 2\int_{a}^{b} x^2 \left(1 + 1\right)^{\frac{1}{2}} dx \\ total_{fuel} &= 2\sqrt{2}\int_{a}^{b} x^2 dx \\ \end{align*}

The rest should be quite straightforward to solve.