Answer:
[tex]\int \int \bigtriangledown F\ X\ dS = \int F\cdot dr[/tex] = 12π
Step-by-step explanation:
The field F is given by:
[tex]F = 3y\hat{i} + y\hat{j}+z\hat{k}[/tex] (1)
The curve C is the ellipse:
[tex]\frac{x^2}{4}+\frac{y^2}{64}=1\\\\(\frac{x}{2})^2+(\frac{y}{8})^2=1[/tex]
In order to calculate the circulation of F around the curve C, you first find the parametric equation for the given ellipse.
The general form of an ellipse equation is:
[tex](\frac{x}{a})^2+(\frac{y}{b})^2=1[/tex]
The parametric equation is:
[tex]r(t)=acost \hat{i} + bsint\hat{j}=2cost\hat{i}+8sint\hat{j}[/tex] (2)
The Stokes's theorem is given by the following identity:
[tex]\int \int \bigtriangledown F\ X\ dS = \int F\cdot dr[/tex]
The path integral is also:
[tex]\int F\cdot dr=\int F(r(t))\cdot dr(t)[/tex] (3)
For F(r(t)) and dr(t) you obtain:
[tex]F(r(t))=3(8sint)\hat{j}+(8sint)\hat{j}+(z)\hat{k}\\\\dr(t)=(-2sint\hat{i}+8cost\hat{j}+0\hat{k})dt\\\\F(r(t))\cdot dr(t)=(-48sin^2t+64cos^2t)dt[/tex]
Next, in the equation (3) you obtain:
[tex]\int_0^{2\pi} (-48sin^2t+64cos^2t)dt=\int_0^{2\pi}(-\frac{48}{2}(1-cos2t)+\frac{64}{2}(1+cos2t))dt\\\\=\int_0^{2\pi}(-24+24cos2t+32+32cos2t)dt\\\\=\int_0^{2\pi}(6+56cos2t)dt\\\\=[6t+56sin2t]_0^{2\pi}=[6(2\pi)-0]=12\pi[/tex]
The circulation of the field around C is 12π
