Question

Use Stokes' Theorem to evaluate the double integral of the curl of F

F(x, y, z) = e^(xy) cos(z)i + x^2 zj + xyk,

S is the hemisphere
x = radical 49 − y^2 − z^2
oriented in the direction of the positive x-axis.

Answer 1

Stokes' theorem relates the surface integral of the curl of
`\vec F` across
`S` to the line integral of
`\vec F` along the boundary of
`S`.

The boundary of
`S` is a circle with radius 7 centered at the origin in the
`x,y`-plane. Parameterize this path by

`\vec r(t) = 7\cos(t)\,\vec\imath + 7\sin(t)\,\vec\jmath`

with
`0\le t\le2\pi`. Observe that
`z=0`, so
`\cos(z) = 1` and the
`\vec\jmath`-component of
`\vec F` contributes nothing. The double integral then reduces to

`\displaystyle \iint_S (\\abla*\vec F)\cdot d\vec S = \int_0^(2\pi) \vec F(\vec r(t)) \cdot (d\vec r)/(dt) \, dt \\\\ ~~~~~~~~ = \int_0^(2\pi) \left(e^(49\cos(t)\sin(t))\,\vec\imath + 49\cos(t)\sin(t)\,\vec\jmath\right) \cdot \left(-7\sin(t)\,\vec\imath + 7\cos(t)\,\vec\jmath\right) \, dt \\\\ ~~~~~~~~ = -7 \int_0^(2\pi) e^(49\cos(t)\sin(t)) \sin(t) \, dt`

Observe that by substituting
`t=u+\pi`, we have

`\sin(t) = \sin(u+\pi) = \sin(u)\cos(\pi) + \cos(u)\sin(\pi) = -\sin(u)`

so that the integral over
`[\pi,2\pi]` can be expressed in terms of the integral over
`[0,\pi]` as

`\displaystyle \int_\pi^(2\pi) e^(49\cos(t)\sin(t)) \sin(t) \, dt = \int_0^\pi -e^(49\cos(t)\sin(t)) \sin(t) \, dt`

Then the integrals over
`[0,\pi]` and
`[\pi,2\pi]` cancel each other and integral of the curl of
`\vec F` is

`\displaystyle -7 \int_0^(2\pi) e^(49\cos(t)\sin(t)) \sin(t) \, dt = -7 \int_0^\pi 0 \, dt = \boxed{0}`