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
`\\abla*\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(\\abla*\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*2\pi \\\\=26\pi`

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

To know more about Integration click the link given below.