# Evaluate the Line integral, where C is the given curve. c xy ds, c: x = t^3, y = t, 0 ≤ t ≤ 3.

This question aims to find the line integral where C is the given curve. An integral is given in the question along with its parameters.

Integration divides the given area, volume, or any other big portion of data into small parts and then finds the summation of these small discrete data. Integration is represented by the symbol of integral.

The integration of some function along the curve in the coordinate axis is called line integral. It is also called the path integral.

Consider the function as:

$f(x,y) = y^3$

\begin{align*}\vec r\left( t \right) & = \left\langle {t^3,t} \right\rangle \\ & \end{align*}

\begin{align*} r’ (t) =\left\langle {3t^2,1} \right\rangle \end{align*}

$ds=|r’(t)|dt$

$ds=\sqrt{(3t^2)^2 + 1^2}dt$

$ds =\sqrt{ (9t^4)+1^2 }dt$

The given integral is $\int y ^ 3 ds$ and integrating this integral with respect to $t$, we get:

$= \int_{ 0 }^{ 3 } f (r (t) )\,ds$

By putting values of $(r(t))$ and $ds$ in the above integral:

$=\int_{ 0 }^{ 3 } t ^ 3 . \sqrt { (9t^4) + 1^2 }\,dt$

Substitute $(9 t ^ 4) + 1 = u$

$9 \times 4t ^ 3 dt + 0 = du$

$t ^ 3 dt = \frac { dt } { 36 }$

$= \int_{0}^{3} t ^ 3 . \sqrt { ( 9t ^ 4 ) + 1 ^ 2 }\, dt$

$=\int_{0}^{3} \sqrt { u } \frac {dt} {36} \$

$=\int_{0}^{3} (\frac {1} {36}) \frac{u^ \frac {3}{2} } { \frac{3}{2}} \ + c$

$=\int_{0}^{3} ( \frac { 1 }{ 54 }) u ^ \frac{3}{2} \ + c$

$= \int_{0}^{3} (\frac {1 } { 54 }) [\sqrt {(9t ^4) + 1 ^2} ] ^ \frac {3}{2}\ + c$

$= (\frac { 1 } { 54 }) [(9 \times 3 ^ { 4 }) + 1] ^ \frac{ 3 }{ 2 } + c – (\frac { 1 }{ 54 }) [(9 \times 0 ^{4} ) + 1] ^ \frac{ 3 }{ 2 } – c$

## Numerical Solution

$= (\frac{1}{54}) [730] ^ \frac{3}{2} – \frac{1}{54}$

$= ( \frac{1}{54}) [730] ^ \frac {3}{2} – 1$

$= 365.28$

The value of the line integral is $365.28$.

## Example

Evaluate $\int 4x^{3}ds$ where $C$ is the line segment from $(-2,-1)$ to $(1,2)$ when $0\leq t \leq 1$.

The line segment is given by the parameterization formulas:

\begin{align*}\vec r\left( t \right) & = \left( {1 – t} \right)\left\langle { – 2, – 1} \right\rangle + t\left\langle {1,2} \right\rangle \\ & = \left\langle { – 2 + 3t, – 1 + 3t} \right\rangle \end{align*}

From the limits:

$x = -2+3t , y = -1+3t$

The line integral by using this path is:

$\int 4x^{3}ds = \int_{1}^{0} 4( -2 + 3t )^3 . \sqrt{9+9}\,dt$

$=12\sqrt{2} (\frac{1}{12}) (-2 + 3t)^4 |_{1}^{0}$

$=12\sqrt{2} (\frac{-5}{4})$

$=-15\sqrt{2}$

$=-21.213$

The value of the line integral is $-21.213$.

Image/Mathematical drawings are created in Geogebra.