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 \(W = 12.56.\)

Answers (1)

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.

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