# Let C be the ellipse contained in the xy plane whose equation is 4x^2 + y^2 = 4, oriented clockwise. The force field F described by

Let C be the ellipse contained in the xy plane whose equation is $4{x}^{2}+{y}^{2}=4$, oriented clockwise. The force field F described by $F\left(x,y,z\right)={x}^{2}i+2xj+{z}^{2}k$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $z=4-4{x}^{2}-{y}^{2},z\ge 0$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $\pi =3.14$, we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore $W=0$.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that $W=12.56.$
c) We can apply Stokes' Theorem and conclude that $W=6.28$
d) We can apply Stokes' Theorem and conclude that $W=12.56.$

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Obiajulu

Step 1
Given the force field,
$F\left(x,y,z\right)={x}^{2}i+2xj+{z}^{2}k$
Also, C is the surface boundary of S:$z=4-4{x}^{2}-{y}^{2},z\ge 0$
Then, C is an ellipse
$4{x}^{2}+{y}^{2}=4$
Then, using the Stoke's Theorem, work done W is given by,
$W={\int }_{C}F\cdot dr$
$=\int {\int }_{S}\left(\mathrm{\nabla }×F\right)\cdot dS$...(1)
Here, $\mathrm{\nabla }×F=\left(0,0,2\right)$
Also, note that the orientation of the surface is clockwise,
Hence,
$n=\left(0,0,-1\right)$
Step 2
Then, from equation(1),
$W=\int {\int }_{S}\left(0,0,2\right)\cdot \left(0,0,-1\right)dS$
$=-2\int {\int }_{S}dS$
$=-2×\left(\pi ×1×2\right)$

Inverting the orientation of the surface S,
$W=\int {\int }_{S}\left(0,0,2\right)\cdot \left(0,0,1\right)dS$
$=-2\int {\int }_{S}dS$
$=-2×\left(\pi ×1×2\right)$

Thus, the correct option is,
b)Inverting the orientation of the surface S, we can apply Stoke's Theorem and conclude that $W=12.56$ units.