It seems that the boundary of [tex]M[/tex] is the circle [tex]x^2+y^2=49[/tex] in the plane [tex]z=0[/tex]. By Stokes' theorem,
[tex]\displaystyle\iint_M(\nabla\times\vec f)\cdot\mathrm d\vec S=\int_{\partial M}\vec f\cdot\mathrm d\vec r[/tex]
Parameterize [tex]\partial M[/tex] by
[tex]\vec r(t)=(7\cos t,7\sin t,0)[/tex]
with [tex]0\le t\le2\pi[/tex]. Then the line integral is
[tex]\displaystyle\int_{\partial M}\vec f(x(t),y(t),z(t))\cdot\mathrm d\vec r=\int_0^{2\pi}(56\sin t,35\cos t,0)\cdot(-7\sin t,7\cos t,0)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{2\pi}(245\cos^2t-392\sin^2t)\,\mathrm dt=\boxed{-147\pi}[/tex]
By applying Stokes theorem, the value of the integral is equal to -52π.
First of all, we would use Stokes theorem to convert the integral as follows:
[tex]\int_s(\Delta \times F).mds = \int_c Fdr[/tex]
where:
c represents the boundary circle on the xy-plane.
Note: The bounding curve (x² + y² = 49) is a circle located at the origin with a radius of 7.
Thus, the parametric equation for the circle is given by:
r(t) = 7costî + 7sintj; t is from 0 to 2π.
Differentiating r(t) with respect to t, we have:
r′(t) = -7sintî + 7costj.
Next, we would compute F(r(t)):
F(r(t)) = 8(7sint)î + 5(7cost)j (z = 0, δz = 0)
By taking the dot product, we have:
F(r(t))⋅r′(t) = -392sin²t + 245cos²t
Note: The double-angle formula for cos is cos(2t) = 2cos²t - 1.
Now, we can integrate from 0 to 2π:
[tex]\int\limits^{2\pi}_{0} (-392sin^2t + 245cos^2t )\, dt \\\\\int\limits^{2\pi}_{0} (\frac{-392}{7} (1-cos2t) + \frac{245}{7}(1+cos2t) )\, dt\\\\\frac{-392}{7}\int\limits^{2\pi}_{0} (1-cos2t) + \frac{245}{7}\int\limits^{2\pi}_{0} (1+cos2t)\\\\\int_s(\Delta \times F).mds = -56[2\pi - 0] + 35[2\pi - 0]\\\\\int_s(\Delta \times F).mds = -112\pi + 70\pi\\\\\int_s(\Delta \times F).mds = -52\pi[/tex]
∬m(∇×f)⋅ds = -52π.
Read more on Stokes theorem here: https://brainly.com/question/13105453
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