\(\int\limits_{C}(x^2+y^2)\,dy\)

This question aims to find the line integral represented by the curve in the given figure.

The anti-derivative of a function is also referred to as the integral of the function. Integration refers to the process of determining a function’s anti-derivative. A family of curves is most commonly represented by an integral of the function.Â More generally, integration refers toÂ adding insignificantly small pieces to determine the content of a continuous region. In calculus, an integral can also be referred to as an area or its generalization. Integration is the process of computing an integralÂ and numerical integration is the approximate computation of an integral.

The surface area in three-dimensional planes is calculated using a line integral. An integral of a function that is usually expressed along a curve in the coordinate system is called a line integral. Moreover, the integrable function may be a scalar or a vector field. Along a curve, we can integrate a scalar or vector-valued function. The value of the line integral can be calculated by adding all of the values of points on the vector field.

## Expert Answer

Given integral is:

$\int\limits_{C}(x^2+y^2)\,dy$

According to the given figure, the above line integral can be split into two parts as:

$\int\limits_{C}(x^2+y^2)\,dy=\int\limits_{C_1}(x^2+y^2)\,dy+\int\limits_{C_2}(x^2+y^2)\,dy$

Where $C$ is the path along the curve $(x^2+y^2)$ from the points $(0,0)$ to $(2,0)$ to $(2,3)$, $C_1$ is the path along the curve from $(0,0)$ to $(2,0)$ and $C_3$ is the path along the curve from $(2,0)$ to $(2,3)$.

Now the equation of $C_1$ through $(0,0)$ to $(2,0)$ is:

$\dfrac{x-0}{2-0}=\dfrac{y-0}{0-0}$

orÂ $y=0$ and so $dy=0$

So, the line integral along $C_1$ becomes:

$\int\limits_{C_1}(x^2+y^2)\,dy=\int\limits_{C_1}(x^2+y^2)\,(0)=0$

And the equation of $C_2$ through $(2,0)$ to $(2,3)$ is:

$\dfrac{x-2}{2-2}=\dfrac{y-0}{3-0}$

orÂ $x=2$

So, the line integral along $C_2$ becomes:

$\int\limits_{C_2}(x^2+y^2)\,dy=\int\limits_{0}^{3}(2^2+y^2)\,dy$

$=\int\limits_{0}^{3}(4+y^2)\,dy$

$=\int\limits_{0}^{3}4\,dy+\int\limits_{0}^{3}y^2\,dy$

$=4[y]_{0}^{3}+\left[\dfrac{y^3}{3}\right]_{0}^{3}$

$=4[3-0]+\dfrac{1}{3}[3^3-0^3]$

$=4[3]+\dfrac{1}{3}[27-0]$

$=12+\dfrac{27}{3}$

$=12+9$

$=21$

## Example

Given $f(x,y)=y+\cos \pi x$ along the line segment $C$ from $(0,2)$ to $(3,4)$. Calculate $\int\limits_{C}f(x,y)\,ds$.

### Solution

First, find the equation of line segment $C$ from $(0,2)$ to $(3,4)$.

The slope-intercept from of the equation of line is given as:

$y=mx+c$

where $m=\dfrac{4-2}{3-0}=\dfrac{2}{3}$

Therefore,Â $y=\dfrac{2}{3}x+c$Â Â Â Â Â Â Â Â Â Â (1)

Now, to find $c$, substitute $(0,2)$ in (1):

$2=\dfrac{2}{3}(0)+c$

$c=2$

So (1) becomes:

$y=\dfrac{2}{3}x+2$

Let $x=t$ then $y=\dfrac{2}{3}t+2$. So, the parametric equations of $C$Â are:

$x(t)=t$ and $y(t)=\dfrac{2}{3}t+2$

Now, $\dfrac{dx}{dt}=1$ and $\dfrac{dy}{dt}=\dfrac{2}{3}$

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

$ds=\sqrt{(1)^2+\left(\dfrac{2}{3}\right)^2}\,dt$

$=\sqrt{1+\dfrac{4}{9}}\,dt$

$=\sqrt{\dfrac{13}{9}}\,dt$

$=\dfrac{\sqrt{13}}{3}\,dt$

And so,Â $\int\limits_{C}f(x,y)\,ds=\int\limits_{0}^{3}\left(\dfrac{2}{3}t+2+\cos \piÂ t\right)\left(\dfrac{\sqrt{13}}{3}\right)\,dt$

$=\dfrac{\sqrt{13}}{3}\int\limits_{0}^{3}\left(\dfrac{2}{3}t+2+\cos \pi Â t\right)\,dt$

$=\dfrac{\sqrt{13}}{3}\left[\dfrac{1}{3}t^2+2t+\dfrac{\sin \pi t}{\pi}\right]_{0}^{3}$

$=\dfrac{\sqrt{13}}{3}\left[\dfrac{1}{3}(3)^2+2(3)+\dfrac{\sin \pi (3)}{\pi}\right]-\dfrac{\sqrt{13}}{3}\left[\dfrac{1}{3}(0)^2+2(0)\dfrac{\sin \pi (0)}{\pi}\right]$

$=\dfrac{\sqrt{13}}{3}[3+6+0]-0$

$=3\sqrt{13}$

*Images/mathematical drawings are created with GeoGebra.*