Question

# 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 $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, 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 $$W = 12.56.$$

2021-02-06

Step 1
Given the force field,
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$
Also, C is the surface boundary of S:$$\displaystyle{z}={4}−{4}{x}^{{2}}−{y}^{{2}},{z}\ge{0}$$
Then, C is an ellipse
$$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$
Then, using the Stoke's Theorem, work done W is given by,
$$\displaystyle{W}=\int_{{C}}{F}\cdot{d}{r}$$
$$\displaystyle=\int\int_{{S}}{\left(\nabla\times{F}\right)}\cdot{d}{S}$$...(1)
Here, $$\displaystyle\nabla\times{F}={\left({0},{0},{2}\right)}$$
Also, note that the orientation of the surface is clockwise,
Hence,
$$n=(0,0,-1)$$
Step 2
Then, from equation(1),
$$\displaystyle{W}=\int\int_{{S}}{\left({0},{0},{2}\right)}\cdot{\left({0},{0},-{1}\right)}{d}{S}$$
$$\displaystyle=-{2}\int\int_{{S}}{d}{S}$$
$$\displaystyle=-{2}\times{\left(\pi\times{1}\times{2}\right)}$$
$$=-12.566\ \text{units}$$
Inverting the orientation of the surface S,
$$\displaystyle{W}=\int\int_{{S}}{\left({0},{0},{2}\right)}\cdot{\left({0},{0},{1}\right)}{d}{S}$$
$$\displaystyle=-{2}\int\int_{{S}}{d}{S}$$
$$\displaystyle=-{2}\times{\left(\pi\times{1}\times{2}\right)}$$
$$=12.566\ \text{units}$$
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.