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Sagot :
Using the Green's Theorem, the one along which the work done by the force is 11π/16.
In the given question we have to find the one along which the work done by the force is the greatest.
The given closed curves in the plane is
[tex]F(x,y)=\left(\frac{x^{2}y}{4} + \frac{y^3}{3}\right)\hat{i}+x\hat{j}[/tex]
Suppose C be a simple smooth closed curve in the plane. It is also oriented counterclockwise.
Let S be the interior of C.
Let P = [tex]\frac{x^{2}y}{4} + \frac{y^3}{3}[/tex] and Q = x
So the partial differentiation is
[tex]\frac{\partial P}{\partial y}=\frac{x^2}{4}+y^2[/tex] and [tex]\frac{\partial Q}{\partial x}[/tex] = 1
By the Green's Theorem, work done by F is given as
W= [tex]\oint \vec{F}d\vec{r}[/tex]
W= [tex]\iint_{S}\left ( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right )dxdy[/tex]
W= [tex]\iint_{S}\left ( 1-\frac{x^2}{y}-y^2 \right )dxdy[/tex]
Let C = x^2+y^2 = 1 and
x = rcosθ, y = rsinθ
0≤r≤1; 0≤θ≤2π
There;
W = [tex]\int_{r=0}^{1}\int_{\theta=0}^{2\pi}\left ( 1-\frac{r^2\cos^2\theta}{4}-r^2\sin^2\theta \right )\left|\frac{\partial(x,y)}{\partial{r,\theta}}\right|d\theta dr[/tex]
and [tex]\frac{\partial (x,y)}{\partial(r, \theta)}=\left|\begin{matrix}\cos\theta &-r\sin\theta \\ \sin\theta & r\cos\theta\end{matrix} \right |[/tex] = r
Thus;
W = [tex]\int_{r=0}^{1}\int_{\theta=0}^{2\pi}\left ( 1-\frac{r^2\cos^2\theta}{4}-r^2\sin^2\theta \right )rd\theta dr[/tex]
After solving
W = 11π/16
Hence, the one along which the work done by the force is 11π/16.
To learn more about Green's Theorem link is here
brainly.com/question/28384298
#SPJ4
The right question is:
Among all simple smooth closed curves in the plane, oriented counterclockwise, find the one along which the work done by the force:
[tex]F(x,y)=\left(\frac{x^{2}y}{4} + \frac{y^3}{3}\right)\hat{i}+x\hat{j}[/tex]
is the greatest. (Hint: First, use Green’s theorem to obtain an area integral—you will get partial credit if you only manage to complete this step.)
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