91Ó°ÊÓ

The given information is $r=\sqrt{x^2+y^2}$.

A line integral is integral in which the function to be integrated is determined along a curve in thecoordinate system.In mathematics and physics, scalar field or scalar-valued function referred to a scalarvalue to everypointin aspace– possiblyphysical space. The scalar may either be a (dimensionless) mathematical number or aphysical quantity.

a) It is known that $r=\sqrt{x^2+y^2}$.

Substitute in the integral with

$$\begin{gathered} x=\cos \mathrm{θ}, \\ y=\sin \mathrm{θ}. \end{gathered}$$

Their derivatives are,

$$\begin{gathered} dx=-\sin \mathrm{θ}d\mathrm{θ}, \\ dy=\cos \mathrm{θ}d\mathrm{θ}. \end{gathered}$$

Find the integral.

$$\begin{gathered} l=∫\left(\cos \mathrm{θ}+2\sin \mathrm{θ}\right)\left(-\sin \mathrm{θ}d\mathrm{θ}\right)-2\cos \mathrm{θ}·\cos \mathrm{θ}d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}\left(-\sin \mathrm{θ}\cos \mathrm{θ}-\left({\sin }^2\mathrm{θ}+{\cos }^2\mathrm{θ}\right)\right)d\mathrm{θ} \\ ={∫}_0^{2\mathrm{π}}\left(\frac{-\sin 2\mathrm{θ}}{2}\right)d\mathrm{θ} \\ ={\left[\frac{\cos 2\mathrm{θ}}{4}-2\mathrm{θ}\right]}_0^{2\mathrm{π}} \\ l=-4\mathrm{π} \end{gathered}$$

b)The path is from to

Substitute in the integral with y = 1.

Its derivative is dx = 0 .

Find the integral.

$$\begin{gathered} l_1={∫}_1^{-1}x+2dx \\ =-4 \end{gathered}$$

The path is from (-1 , 1) to (-1 ,-1)

Substitute in the integral with x = -1.

Its derivative is dx = 0.

Find the integral.

$$\begin{gathered} l_2={∫}_1^{-1}2dy \\ =-4 \end{gathered}$$

The path is from (-1 , 1) to (-1 , 1) .

Substitute in the integral with y = -1.

Its derivative is dy = 0.

Find the integral.

$$\begin{gathered} l_3={∫}_{-1}^1\left(x-2\right)dx \\ =-4 \end{gathered}$$

Following is the total integral:

$$\begin{gathered} l=l_1+l_2+l_3+l_4 \\ =-4-4-4-4 \\ =-16 \end{gathered}$$

(c) The path is from (0,1) to (-1,0).

Find the equation of line from point (0,1) to (-1,0).

$$\begin{gathered} \frac{y-1}{x-0}=\frac{0-1}{-1-0} \\ y=x+1 \end{gathered}$$

Substitute in the integral with y = x + 1.

Its derivative is dy = dx.

$$\begin{gathered} l_1={∫}_0^{-1}\left(x+2x+2-2x\right)dx \\ =\frac{-3}{2} \end{gathered}$$

The path is from (-1,0) to (0,-1)

Find the equation of line from point (-1,0) to (0,-1)

$$\begin{gathered} \frac{y-0}{x+1}=\frac{-1-0}{0+1} \\ y=-x-1 \end{gathered}$$

Substitute in the integral with y = - x - 1.

Its derivative is dy = -dx.

$$\begin{gathered} l_2={∫}_{-1}^0\left(x+-2x-2+2x\right)dx \\ =\frac{-5}{2} \end{gathered}$$

The path is from (0, -1) to (1,0)

Find the equation of line from point ( -1,0) to (0, -1)

$$\begin{gathered} \frac{y+1}{x}=\frac{1}{1} \\ y=x-1 \end{gathered}$$

Substitute in the integral with y = x - 1.

Its derivative is dy = dx.

$$\begin{gathered} l_3={∫}_0^1\left(x+2x-2-2x\right)dx \\ =\frac{-3}{2} \end{gathered}$$

The path is from (1,0) to (1,0)

Find the equation of line from point (1,0) to (1,0)

$$\begin{gathered} \frac{y-0}{x-1}=\frac{1}{-1} \\ y=1-x \end{gathered}$$

Substitute in the integral with.

Its derivative is.

$$\begin{gathered} l_4={∫}_1^0\left(x+2x+2-2x\right)dx \\ =\frac{-5}{2} \end{gathered}$$

Find the total integral.

$$\begin{gathered} l=l_1+l_2+l_3+l_4 \\ =-8 \end{gathered}$$

Hence, the solution to this problem is mentioned below.

$$\begin{gathered} a) \\ b) \\ c) \end{gathered}$$$$\begin{gathered} \mathrm{l}=4\mathrm{Ï€} \\ \mathrm{l}=-16 \\ \mathrm{l}=-8 \end{gathered}$$