# 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.

## Expert Answer

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:

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

Images/Mathematical drawings are created with GeoGebra.

5/5 - (6 votes)