${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$, where C is the curve on the paraboloid $z=x^2+y^2$that lies above the unit circle, traversed counterclockwise with respect to the outwards-pointing normal vector, and where

$\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$

$\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$

The goal is to calculate the line integral.localid="1650767632481" ${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$, where the curve C is defined as follows:

The curve C is the paraboloid curve.$z=x^2+y^2$and that lies above the unit circle, traversed counterclockwise with respect to the normal vector pointing outward.

To calculate this integral, use Stokes' Theorem. According to Stokes' Theorem,

"Assume that S is an oriented, smooth or piecewise-smooth surface bounded by a curve C.. Assume is an oriented unit normal vector of S and C with a parametrization that traverses C counterclockwise with respect to n.

If a vector field $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined on S, then,

localid="1650767639116" ${\int }_C \mathbf{F}(x,y,z)\times d\mathbf{r}={\iint }_S \mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}dS.$

First find the curl of the vector field $\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$

$=\left[\frac{\mathrm{\partial }{F}_3}{\mathrm{\partial }y}-\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }z}\right]\mathbf{i}-\left[\frac{\mathrm{\partial }{F}_3}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }z}\right]\mathbf{j}+\left[\frac{\mathrm{\partial }{F}_2}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{F}_1}{\mathrm{\partial }y}\right]\mathbf{k}$first, determine the vector field's curl.

$\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$

A vector field's curl $\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}$is defined as follows:

$\mathrm{curl}\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }x}& \frac{\mathrm{\partial }}{\mathrm{\partial }y}& \frac{\mathrm{\partial }}{\mathrm{\partial }z}\\ F_1(x,y,z)& F_2(x,y,z)& F_3(x,y,z)\end{array}\right|$

Then, A vector field's curl$\mathbf{F}(x,y,z)=(3x+y-z)\mathbf{i}+(4y-2z)\mathbf{j}+(x-3z)\mathbf{k}$will be,

$\mathrm{curl}\mathbf{F}(x,y,z)=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ \frac{\mathrm{\partial }}{\mathrm{\partial }x}& \frac{\mathrm{\partial }}{\mathrm{\partial }y}& \frac{\mathrm{\partial }}{\mathrm{\partial }z}\\ (3x+y-z)& (4y-2z)& (x-3z)\end{array}\right|$

$\begin{array}{r}=\left[\frac{\mathrm{\partial }(x-3z)}{\mathrm{\partial }y}-\frac{\mathrm{\partial }(4y-2z)}{\mathrm{\partial }z}\right]\mathbf{i}-\left[\frac{\mathrm{\partial }(x-3z)}{\mathrm{\partial }x}-\frac{\mathrm{\partial }(3x+y-z)}{\mathrm{\partial }z}\right]\mathbf{j}\\ \end{array}$

$+\left[\frac{\mathrm{\partial }(4y-2z)}{\mathrm{\partial }x}-\frac{\mathrm{\partial }(3x+y-z)}{\mathrm{\partial }y}\right]\mathbf{k}$

localid="1650767650444" $=0-(-2)]\mathbf{i}-1-(-1\left)\right]\mathbf{j}+[0-1]k$

localid="1650766213891" $=2i+2j+2k$

localid="1650766235827" $<2,-2,1>$

The letter n should be pointing outwards.

If $z=z(x,y)$, then the following vector;

is the normal to the surface

Now, for $z=x^2+y^2$, It produces the perpendicular vector to this surface;

$=⟨-2x,-2y,1⟩$

Then, the value of $curl\mathbf{F}(x,y,z)·\mathbf{n}$will be,

localid="1650766271635" $\mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}=⟨2,-2,-1⟩\times ⟨-2x,-2y,1⟩$

localid="1650767664870" $$\begin{gathered} =\left(2\right)\times (-2x)+(-2)\times (-2y)+(-1)\times \left(1\right) \\ -4x+4y-1 \end{gathered}$$

The region of integration D is the unit disk in the x y-plane because the curve C lies above the unit circle when traversed counterclockwise.

The region of integration is described in polar coordinates as follows.

$D=\left\{\right(r,\theta )\mid 0\le r\le 1,0\le \theta \le 2\pi \}.$

In this case

$x=r\cos \theta ,y=r\sin \theta ,x^2+y^2=r^2$

And

$dA=rdrd\theta$

Now, apply Stokes' Theorem (1) to the integral.${\int }_C\mathbf{F}(x,y,z)\times d\mathbf{r}$as follows:

${\int }_C \mathbf{F}(x,y,z)\times d\mathbf{r}={\iint }_S \mathrm{curl}\mathbf{F}(x,y,z)\times \mathbf{n}dS$

$={\iint }_D (-4x+4y-1)dA$

$={\int }_0^{2\pi } {\int }_0^1 (-4r\cos \theta +4r\sin \theta -1)rdrd\theta$

$={\int }_0^{2\pi } {\int }_0^1 \left(\right(\sin \theta -\cos \theta )4r-1)rdrd\theta$

$={\int }_0^{2\pi } {\int }_0^1 \left((\sin \theta -\cos \theta )4r^2-r\right)drd\theta$

$={\int }_0^{2\pi } \left[{\int }_0^1 \left((\sin \theta -\cos \theta )4r^2-r\right)dr\right]d\theta$

$={\int }_0^{2\pi } {\left[(\sin \theta -\cos \theta )\frac{4r^3}{3}-\frac{r^2}{2}\right]}_0^{1}d\theta$

localid="1650766301574" $={\int }_0^4 \left[\left((\sin \theta -\cos \theta )\frac{4\times 1^3}{3}-\frac{1^2}{2}\right)-\left((\sin \theta -\cos \theta )\frac{4\times 0^3}{3}-\frac{0^2}{2}\right)\right]d\theta$

$={\int }_0^{2\pi } \left((\sin \theta -\cos \theta )\frac{4}{3}-\frac{1}{2}\right)d\theta$

$={\int }_0^{2\pi } \left(\frac{4}{3}\sin \theta -\frac{4}{3}\cos \theta -\frac{1}{2}\right)d\theta$

$={\left[-\frac{4}{3}\cos \theta +\frac{4}{3}\sin \theta -\frac{1}{2}\theta \right]}_0^{2\pi }$localid="1650766362431" $=\left(-\frac{4}{3}\cos 2\pi +\frac{4}{3}\sin 2\pi -\frac{1}{2}\times 2\pi \right)-\left(-\frac{4}{3}\cos 0+\frac{4}{3}\sin 0-\frac{1}{2}\times 0\right)$

localid="1650766338659" $$\begin{gathered} =\left(-\frac{4}{3}\times 1+\frac{4}{3}\times 0-\frac{1}{2}\times 2\pi \right)-\left(-\frac{4}{3}\times 1+\frac{4}{3}\times 0-\frac{1}{2}\times 0\right) \\ =-\mathrm{\pi } \end{gathered}$$