Question

(1 point) Use Stokes' theorem to evaluate (V x F). dS where F(x, y, z) = -13yzi + 13xzj +4(x2 + y2)zk and S is the part of the paraboloid z = x2 + y2 that lies inside the cylinder x2 + y2 = 1, oriented upward. JS

Answer 1

Stokes' theorem equates the integral of $\nabla\times\vec F(x,y,z)$ across $S$ with the integral of $\vec F(x,y,z)$ along the boundary of $S$, which is the intersection of the paraboloid $z=x^2+y^2$ and the cylinder $x^2+y^2=1$, or the circle in the plane $z=1$ centered at (0, 0, 1) with radius 1.

Parameterize this intersection (call it $C$) by

$\vec r(t)=\cos t\,\vec\imath+\sin t\,\vec\jmath+\vec k$

with $0\le t\le2\pi$. Then

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r$

$=\displaystyle\int_0^{2\pi}(-13\sin t\,\vec\imath+13\cos t\,\vec\jmath+4\,\vec k)\cdot(-\sin t\,\vec\imath+\cos t\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}(13\sin^2t+13\cos^2t)\,\mathrm dt$

$=\displaystyle13\int_0^{2\pi}\mathrm dt=\boxed{26\pi}$

Answer 2

By using Stokes's theorem to the required (VxF).dS  is 26π.

Given that,

Function F(x, y, z) = -13yzi + 13xzj +4(x2 + y2)zk .

And S is the part of the paraboloid z = x2 + y2 that lies inside the cylinder x2 + y2 = 1, oriented upward. JS.

We have to determine,

Use Stokes' theorem to evaluate (V x F).dS.

According to the question,

By using Stokes's theorem to evaluate (V x F).dS .

F(x, y, z) = -13yzi + 13xzj +4(x2 + y2)zk .

Stokes' theorem says the integral of the curl of over F is s equal to the integral of F along the boundary of S , with counterclockwise orientation (when viewed from above).

This boundary is the circle $x^{2} + y^{2} =1$ set in the plane $z= 1$.

Then,

The parameterize path is given by the,

$\vec{r} = cost\vec{i} + sint \vec{j} +\vec{k} \ \ with\ 0\leq t\leq 2\pi$

Then,

$\int \int_s \bigtriangledown . \vec{F} . \vec{ds} =\int_c \vec{F}.\vec{dr}$

$= \int^{2\pi }_0 (-13sint\vec{i} + 13cost\vec{j} +4\vec{k} ) . (-sint\vec{k} + cost\vec{j}).dt\\\\= \int^{2\pi }_0 (13sin^2t +13cos^2t)dt\\\\= 13\int^{2\pi }_0 1.dt \\\\=13 [{t}]^{2\pi }_0\\\\= 13 [ 2\pi -0] \\\\= 13\times2\pi \\\\=26\pi$

Hence, By using Stokes's theorem to the required (VxF).dS  is 26π.

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