 # Evaluate the line integral, where C is the given curve $$\int\limits_{C}xy\,ds$$. $$C:x=t^2,\,\,y=2t,\,\,0\leq t\leq 5$$.

This question aims to find the given line integral using the parametric equations of the curve $C$.

A line integral represents the integration of a function along a curve. It can also be regarded as a path integral, curvilinear integral, or curve integral.

The line integrals are the extension of simple integrals (which helps in finding areas of flat and two-dimensional surfaces) and can be used to find the areas of the surfaces that curve out into three dimensions. It is integral that integrates a function along a curve in the coordinate system.

The function to be integrated can be defined as either a scalar or a vector field. Along a curve, we can integrate both scalar and vector-valued functions. The vector line integral can be calculated by adding the values of all the points on the vector field.

Since, $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$

Therefore,  $\dfrac{dx}{dt}=2t$ and $\dfrac{dy}{dt}=2$

So, $ds=\sqrt{(2t)^2+\left(2\right)^2}\,dt$

$=\sqrt{4t^2+4}\,dt$

$=2\sqrt{t^2+1}\,dt$

And $\int\limits_{C}xy\,ds$ $=\int\limits_{0}^{5}(t^2)(2t)(2\sqrt{t^2+1})\,dt$

$=4\int\limits_{0}^{5} t^3\sqrt{1+t^2}\,dt$

Or, $\int\limits_{C}xy\,ds=2\int\limits_{0}^{5} t^2\sqrt{1+t^2}\cdot 2t\,dt$

Applying integration by substitution, let:

$1+t^2=u\implies t^2=u-1$

and $du=2t\,dt$

Also, when $t=0$, $u=1$

and when $t=5$, $u=26$

Therefore,  $\int\limits_{C}xy\,ds=2\int\limits_{1}^{26} (u-1)\sqrt{u}\,du$

$=2\int\limits_{1}^{26} (u^{3/2}-u^{1/2})\,du$

$=2\left[\dfrac{u^{5/2}}{5/2}-\dfrac{u^{3/2}}{3/2}\right]_{1}^{26}$

$=4\left[\dfrac{u^{5/2}}{5}-\dfrac{u^{3/2}}{3}\right]_{1}^{26}$

$=4\left[\dfrac{(26)^{5/2}-(1)^{5/2}}{5}-\dfrac{(26)^{3/2}-(1)^{3/2}}{3}\right]$

$=4\left[\dfrac{(26)^2\sqrt{26}-1}{5}-\dfrac{26\sqrt{26}-1}{3}\right]$

$=4\left[\dfrac{676\sqrt{26}}{5}-\dfrac{1}{5}-\dfrac{26\sqrt{26}}{3}+\dfrac{1}{3}\right]$

$=4\left[\dfrac{(2028-130)\sqrt{26}}{15}+\dfrac{5-3}{15}\right]$

$\int\limits_{C}xy\,ds=\dfrac{4}{15}[1898\sqrt{26}+2]$ Graph of the given curve along with its surface area

## Example 1

Determine the line integral $\int\limits_{C}\left(\dfrac{y}{1+x^2}\right)\,ds$, where $C$ is a curve given by the parametric equations: $x=t,\,y=2+t$ for $0\leq t\leq 1$.

### Solution

Since, $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$

Therefore, $\dfrac{dx}{dt}=1$ and $\dfrac{dy}{dt}=1$

So, $ds=\sqrt{(1)^2+\left(1\right)^2}\,dt$

$=\sqrt{1+1}\,dt$

$=\sqrt{2}\,dt$

And $\int\limits_{C}\left(\dfrac{y}{1+x^2}\right)\,ds$ $=\int\limits_{0}^{1}\left(\dfrac{2+t}{1+t^2}\right)(\sqrt{2})\,dt$

$=\sqrt{2}\int\limits_{0}^{1} \left(\dfrac{2}{1+t^2}+\dfrac{t}{1+t^2}\right)\,dt$

$=\sqrt{2}\left[\int\limits_{0}^{1} \dfrac{2}{1+t^2}\,dt+\int\limits_{0}^{1} \dfrac{t}{1+t^2}\,dt\right]$

$=\sqrt{2}\left[2\tan^{-1}(t)+\dfrac{\ln(1+t^2)}{2}\right]_{0}^{1}$

Applying the limits of integration as:

$=\sqrt{2}\left(2\tan^{-1}(1)+\dfrac{\ln(1+(1)^2)}{2}\right)-\sqrt{2}\left(2\tan^{-1}(0)+\dfrac{\ln(1+(0)^2)}{2}\right)$

$=\sqrt{2}\left(2\cdot \dfrac{\pi}{4}+\dfrac{\ln(2)}{2}\right)-\sqrt{2}\left(0+0\right)$

$=\sqrt{2}\left(\dfrac{\pi}{2}+\dfrac{\ln(2)}{2}\right)$

$=\sqrt{2}\left(\dfrac{\pi+\ln(2)}{2}\right)$

Or $\int\limits_{C}\left(\dfrac{y}{1+x^2}\right)\,ds$ $=\dfrac{\pi+\ln(2)}{\sqrt{2}}$

## Example 2

Work out the line integral $\int\limits_{C}xy\,ds$, where $C$ is a curve defined by the parametric equations: $x=\cos t,\,y=\sin t$ for $0\leq t\leq \pi$.

### Solution

Since, $ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt$

Therefore,  $\dfrac{dx}{dt}=-\sin t$ and $\dfrac{dy}{dt}=\cos t$

So, $ds=\sqrt{(-\sin t)^2+\left(\cos t\right)^2}\,dt$

$=\sqrt{\sin^2t+\cos^2t}\,dt$

$=\sqrt{1}\,dt$

So, $ds=1\cdot dt$

And $\int\limits_{C}xy\,ds$ $=\int\limits_{0}^{\pi}(\cos t)(\sin t)(1)\,dt$

$=\int\limits_{0}^{\pi} \cos t\sin t\,dt$

$=\int\limits_{0}^{\pi} \sin t (\cos t\,dt)$

Now, by using the power rule:

$=\left[\dfrac{\sin^2 t}{2}\right]_{0}^{\pi}$

Applying the limits of integration as:

$=\left[\dfrac{\sin^2 (\pi)}{2}-\dfrac{\sin^2 (0)}{2}\right]$

$=\left[\dfrac{0}{2}-\dfrac{0}{2}\right]$

Or $\int\limits_{C}xy\,ds=0$

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