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Stokes' theorem relates the surface integral of the curl of [tex]\vec F[/tex] across [tex]S[/tex] to the line integral of [tex]\vec F[/tex] along the boundary of [tex]S[/tex].
The boundary of [tex]S[/tex] is a circle with radius 7 centered at the origin in the [tex]x,y[/tex]-plane. Parameterize this path by
[tex]\vec r(t) = 7\cos(t)\,\vec\imath + 7\sin(t)\,\vec\jmath[/tex]
with [tex]0\le t\le2\pi[/tex]. Observe that [tex]z=0[/tex], so [tex]\cos(z) = 1[/tex] and the [tex]\vec\jmath[/tex]-component of [tex]\vec F[/tex] contributes nothing. The double integral then reduces to
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec S = \int_0^{2\pi} \vec F(\vec r(t)) \cdot \frac{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[/tex]
Observe that by substituting [tex]t=u+\pi[/tex], we have
[tex]\sin(t) = \sin(u+\pi) = \sin(u)\cos(\pi) + \cos(u)\sin(\pi) = -\sin(u)[/tex]
so that the integral over [tex][\pi,2\pi][/tex] can be expressed in terms of the integral over [tex][0,\pi][/tex] as
[tex]\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[/tex]
Then the integrals over [tex][0,\pi][/tex] and [tex][\pi,2\pi][/tex] cancel each other and integral of the curl of [tex]\vec F[/tex] is
[tex]\displaystyle -7 \int_0^{2\pi} e^{49\cos(t)\sin(t)} \sin(t) \, dt = -7 \int_0^\pi 0 \, dt = \boxed{0}[/tex]