Use Stokes’ Theorem to evaluate S curl F · dS. F(x, y, z) = 4y cos z i + ex sin z j + xey k, S is the hemisphere x2 + y2 + z2 = 49, z ≥ 0, o

Question

Use Stokes’ Theorem to evaluate S curl F · dS. F(x, y, z) = 4y cos z i + ex sin z j + xey k, S is the hemisphere x2 + y2 + z2 = 49, z ≥ 0, oriented upward.

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Evelyn 18 mins 2021-09-13T11:00:02+00:00 1 Answer 0

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    2021-09-13T11:01:47+00:00

    Answer:

    -196π

    Step-by-step explanation:

    F can be rewritten as

    \bf F(x,y,z)=(4ycos(z),e^xsin(z),xe^y)

    if S is the upper hemisphere  

    \bf x^2 + y^2 + z^2 = 49

    oriented upward, then the border of S is the circle C

    \bf x^2 + y^2 = 7^2

    traversed counterclockwise

    By Stoke’s theorem

    \bf \displaystyle\iint_S(curl\;F)dS=\displaystyle\int_{C}F.dC

    where C is the circle of center (0,0,0) and radius 7 on the XY-plane  traveled counterclockwise.

    This circle can be parametrized as

    r(t) = (7cos(t), 7sin(t),0) with 0 ≤t ≤ 2π  

    Computing the curve integral

    \bf \displaystyle\int_{C}F.dC=\displaystyle\int_{0}^{2\pi}F(r(t))\bullet r'(t)dt=\displaystyle\int_{0}^{2\pi}F(7cos(t),7sin(t),0)\bullet (-7sin(t),7cos(t),0)dt=\\\\=\displaystyle\int_{0}^{2\pi} (28sin(t),0,7cos(t)e^{7sin(t)})\bullet(-7sin(t),7cos(t),0)dt=\\\\-196\displaystyle\int_{0}^{2\pi}sin^2(t)dt=\boxed{-196\pi}

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