Question

Use Stokes' Theorem to calculate the circulation of the field F around the curve C in the indicated direction. F = 3yi + yj+zk

C: counterclockwise path around the boundary of the ellipse x2/4+y2/64=1

Answer 1

`\int \int \bigtriangledown F\ X\ dS = \int F\cdot dr` = 12π

Explanation:

The field F is given by:

`F = 3y\hat{i} + y\hat{j}+z\hat{k}` (1)

The curve C is the ellipse:

`(x^2)/(4)+(y^2)/(64)=1\\\\((x)/(2))^2+((y)/(8))^2=1`

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:

`((x)/(a))^2+((y)/(b))^2=1`

The parametric equation is:

`r(t)=acost \hat{i} + bsint\hat{j}=2cost\hat{i}+8sint\hat{j}` (2)

The Stokes's theorem is given by the following identity:

`\int \int \bigtriangledown F\ X\ dS = \int F\cdot dr`

The path integral is also:

`\int F\cdot dr=\int F(r(t))\cdot dr(t)` (3)

For F(r(t)) and dr(t) you obtain:

`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`

Next, in the equation (3) you obtain:

`\int_0^(2\pi) (-48sin^2t+64cos^2t)dt=\int_0^(2\pi)(-(48)/(2)(1-cos2t)+(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`

The circulation of the field around C is 12π