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The given information is ${鈭�}_c{y}^{2}dx+2xdy+dz$.${鈭�}_c{y}^{2}dx+2xdy+dz$.

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 scalarvalueto everypointin aspace鈥� possiblyphysical space. The scalar may either be a (dimensionless)mathematical numberor aphysical quantity.

a) In order to get from the origin to the point (1,1,1) take an infinite number of routes.

One single route from the origin to the point (1,0,0) on a straight line. In this segment the variable x changes continuously from 0 to 1, but the other two variables remain 0. This means that the line integral is zero.

y=z

=dy

=dz

=0

The second displacement is from the point (1,0,0) to the point (1,0,1) in this segment z changes continuously from 0 to 1 , but the variable x is fixed to the value 1 , which means

dx=0,

y=dy

=0

This means that the last contribution is not zero.

$$\begin{gathered} {鈭�}_{c}dz={鈭�}_0^{1}dz \\ =1 \end{gathered}$$

The last segment is from (1,0,1) to (1,1,1) in this segment the variables x and z are fixed at the value 1 , which gives,

dz=dx

=0

The variable y changes continuously from 0 to 1.

$$\begin{gathered} {鈭�}_{c}2xdy={鈭�}_0^{1}2.1dz \\ =2 \end{gathered}$$

The total line integral is ${鈭�}_c{y}^{2}dx+2xdy+dz=3.$

b)Take the curve C to go from the origin to the point (1,1,0) as a circle.

$x^2+y^2-2y=0$ 鈥︹赌�.(1)

The equation (1) can be written in different form.

$x^2+{\left(y-1\right)}^2-1=0$

This is a circle with radius 1 and center at y=1 .

Write the coordinates in trigonometric form.

$$\begin{gathered} x=\cos 胃, \\ y-1=\sin 胃. \end{gathered}$$

Simplify the coordinates.

$$\begin{gathered} y=1+\sin 胃 \\ {y}^2=1+2\sin 胃+{\sin }^2胃 \\ dx=-\sin 胃d胃 \\ dy=\cos 胃d胃 \end{gathered}$$

Here, $胃$changes from $\frac{3蟺}{2}$ to $2蟺$.

Evaluate the integral.

role="math" localid="1659172605940" $$\begin{gathered} {鈭�}_{\frac{3蟺}{2}}^{2蟺}d胃\left(1+2\sin 胃+{\sin }^2胃\right)\left(-\sin 胃\right)+2\cos 胃\left(\cos 胃\right)={鈭�}_{\frac{3蟺}{2}}^{2蟺}\left(-\sin 胃-{\sin }^3胃+2\cos \left(2胃\right)\right)d胃 \\ =\cos {\overline{)胃}}_{\frac{3蟺}{2}}^{2蟺}-{\overline{)\frac{1}{12}\overline{)\left(\cos \left(3胃\right)-9\cos \left(x\right)\right)}}}_{\frac{3蟺}{2}}^{2蟺}{\overline{)+\frac{2}{2}\sin \left(2胃\right)}}_{\frac{3蟺}{2}}^{2蟺} \\ =1-\frac{1}{12}\left(1-9\right) \\ =\frac{5}{3} \end{gathered}$$

The last segment is a straight line from (1,1,0) to (1,1,1) where z changes continuously from 0 to 1 , and x and y are fixed at a value 1 , which gives,

$$\begin{gathered} dx=dy \\ =0 \end{gathered}$$

Therefore, the last segment is equal to 1 .

$$\begin{gathered} {鈭�}_c{y}^{2}dx+2xdy+dz=\frac{5}{3}+1 \\ =\frac{8}{3} \end{gathered}$$

Hence,the solution to this problem is mentioned below.

a) 3 ,

b) $\frac{8}{3}$.