# Evaluate the line integral where c is the given curve.

$\boldsymbol{ \oint xy \ ds \text{ where s is defined by } x = t^2 \text{ and } y = 2t \text{ over the interval } 0 \leq t \leq 4 }$

The aim of this question is to learn how to solve line integrals over some closed surfaces.

To solve this question, we simply find the value of the $ds$ using the following formula:

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

And then solve the integral after applying the given constraints.

Given:

$x = t^2 \Rightarrow \dfrac{ dx }{ dt } = 2t$

$x = 2t \Rightarrow \dfrac{ dy }{ dt } = 2$

Evaluating $ds$:

$ds = \sqrt{ ( 2t )^2 + ( 2 )^2 } dt = \sqrt{ 4t^2 + 4 } dt$

$ds = \sqrt{ 4 (t^2 + 1) } dt = 2 \sqrt{ t^2 + 1 } dt$

Applying all the constraints to the line integral:

$\int xy \ ds = \int_{t=0}^{t=4} (t^2)(2t)(2 \sqrt{ t^2 + 1 })dt$

$\int xy \ ds = 4 \int_{t=0}^{t=4} (t^2)(\sqrt{ t^2 + 1 })(t)dt \ ……………. \ (1)$

Let’s assume:

$t^2 + 1 = u^2 \Rightarrow 2tdt = 2udu \Rightarrow tdt = udu$

Which means:

$u = \sqrt{ t^2 + 1 }$

So:

$t = 0 \rightarrow u = \sqrt{ (0)^2 + 1 } = 1$

$t = 4 \rightarrow u = \sqrt{ (4)^2 + 1 } = \sqrt{ 17 }$

Substituting these values in equation (1):

$\int xy \ ds = 4 \int_{u=1}^{u=\sqrt{ 17 }} (u^2 -1 )(\sqrt{ u^2 })udu$

$\int xy \ ds = 4 \int_{u=1}^{u=\sqrt{ 17 }} (u^2 -1 )u^2du$

$\int xy \ ds = 4 \int_{u=1}^{u=\sqrt{ 17 }} (u^4 -u^2)du$

$\int xy \ ds = 4 \bigg | \dfrac{u^5}{5} – \dfrac{u^3}{3} \bigg |_{u=1}^{u=\sqrt{ 17 }}$

$\int xy \ ds = \dfrac{ 4 }{ 15 }\bigg | 3u^5 – 5u^3 \bigg |_{u=1}^{u=\sqrt{ 17 }$

$\int xy \ ds = \dfrac{ 4 }{ 15 }\bigg ( 3(\sqrt{ 17 })^5 – 5(\sqrt{ 17 })^3 – 3(1)^5 + 5(1)^3 \bigg )$

$\int xy \ ds = \dfrac{ 4 }{ 15 }\bigg ( 3574.73 – 350.46 – 3 + 5 \bigg )$

$\int xy \ ds = \dfrac{ 4 }{ 15 } 3225.27$

$\int xy \ ds = 860.33$

## Numerical Result

$\int xy \ ds = 860.33$

## Example

Calculate the value of the following line integral as per the given constraints:

$\boldsymbol{ \oint xy \ ds \text{ where s is defined by } x = 4t \text{ and } y = 3t \text{ over the interval } 0 \leq t \leq 4 }$

Here:

$\dfrac{ dx }{ dt } = 4, \ \dfrac{ dy }{ dt } = 3$

So:

$ds = \sqrt{ ( 4 )^2 + ( 3 )^2 } dt = \sqrt{ 16 + 9 } dt = \sqrt{ 25 } dt = 5 dt$

Applying all the constraints to the line integral:

$\int xy \ ds = \int_{t=0}^{t=4} (4t)(3t)(5) dt = \int_{t=0}^{t=4} 60 t^2 dt$

$\int xy \ ds = \bigg | \dfrac{60 t^3}{3} \bigg |_{0}^{4} = \dfrac{60 (4)^3}{3} – \dfrac{60 (0)^3}{3})$

$\int xy \ ds = 1280$