# Evaluate the line integral, where c is the given curve. $\int_{c} xy ds$, c : x = t^2, y = 2t, 0 ≤ t ≤ 2.

The motivation of this question is to find the line integral. A line integral is an integral of a function along a path or curve, and a curve in the XY-plane works with two variables.

To understand this topic, knowledge of curves and straight lines in geometry is required. Techniques of integration and differentiation need calculation.

The curve is given in parametric form, so the formula is:

$ds = \int_{t_1}^{t_2} \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2}$

Given as:

$x = t^{2}, \hspace{0.4in} y = 2t$

$\dfrac{dx}{dt} = 2t, \hspace{0.4in} \dfrac{dy}{dt} = 2$

$ds = \int_{0}^{2} \sqrt{(2t)^2 + (2)^2} \, dt$

$ds = 2\int_{0}^{2} \sqrt{t^{2} + 1}dt$

Substituting the given values, we get:

$t = \tan{\theta} \implies \hspace{0.4in} dt = sec^{}\theta$

$At \hspace{0.2in} t= 0; \hspace{0.2in} \theta = 0$

$At \hspace{0.2in} t = 2; \hspace{0.2in} \tan{\theta} = 2 \implies \theta = \tan^{-1}(2) = 1.1$

We get:

$ds = 2\int_{0}^{1.1} \sqrt{1 + tan^{2}} \sec^{2}{\theta} \,d{\theta}$

$ds = 2\int_{0}^{1.1} \sec^{3}{\theta} d{\theta}$

$ds = 2\int_{0}^{1.1} \sec{\theta} \sec^{2}{\theta} {d{\theta}}$

Now, Integration by parts, taking $\sec\theta$ as first function

$I = 2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} – \int_{0}^{1.1} \tan \theta\bigg(\frac{d}{d \theta} \sec \theta\bigg) d \theta \bigg]$

$I =2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} – \int_{0}^{1.1}\tan^{2} \theta \sec \theta d \theta \bigg]$

$I =2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} – \int_{0}^{1.1}(\sec^{2}\theta-1) \sec \theta d \theta\bigg]$

$I =2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} – \int_{0}^{1.1}\sec^{3} \theta d \theta+\int_{0}^{1.1} \sec \theta d \theta\bigg]$

$I =2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} – I + \int_{0}^{1.1}\sec \theta d \theta \bigg]$

$I + I =2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} + \int_{0}^{1.1}\sec \theta d \theta \bigg]$

$2 I =2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} + \int_{0}^{1.1}\sec \theta d\theta \bigg]$

$2 I =2 \bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} + ln|\sec \theta + \tan \theta|_0^{1.1}\bigg]$

$I =\bigg[[\sec \theta{\tan \theta}\big]_0^{1.1} + ln|\sec \theta + \tan \theta|_0^{1.1}\bigg]$

Since:

$\tan\theta = x = \frac{P}{B}$

$\sin\theta = \frac{x}{\sqrt{(1 + x^{2})}}$

$\cos\theta = \frac{1}{\sqrt{(1 + x^{2})}}$

## Numerical Result

The above trigonometric ratios are obtained by using Pythagoras Theorem.

$ds = [x\sqrt{(1 + x^{2})}]_0^{1.1} + ln|x + \sqrt{(1 + x^{2})}|_0^{1.1}$

$ds = [1.1 \sqrt{(1 + (1.1)^{2}}) – 0] + [ln|1.1 + \sqrt{1 + (1.1)^{2}}| – ln|1|]$

$ds = 3.243$

## Example:

Given the curve $C:$ $x^2/2 + y^2/2 =1$, find the line integral.

$\underset{C}{\int} xy \, ds$

The curve is given as:

$\dfrac{x^2}{2} + \dfrac{y^2}{2} = 1$

The equation of the ellipse in parametric form is given as:

$x = a \cos t, \hspace{0.2in} y = b \sin t, \hspace{0.4in} 0 \leq t \leq \pi/2$

The line integral becomes:

$I = \underset{C}{\int} xy \, ds$

$I = \int_{0}^{\frac{\pi}{2}} a \cos t.b \sin t \sqrt{(-a \sin t)^2 + (b \cos t)^2} \, dt$

Solving the integral, we get:

$I = \dfrac{ab (a^2 + ab + b^2)}{3(a + b)}$

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