Let F= 6(x+y)i + 6 sin(y)j,. Find the line integral of F around the perimeter of the rectangle with corners (5,0),(5,7),(-3,7),(-3,0) traver

Question

Let F= 6(x+y)i + 6 sin(y)j,. Find the line integral of F around the perimeter of the rectangle with corners (5,0),(5,7),(-3,7),(-3,0) traversed in that order.

line integral =

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Arianna 2 weeks 2021-09-13T16:23:41+00:00 1 Answer 0

Answers ( )

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    2021-09-13T16:25:07+00:00

    Answer:

    333.04656

    Step-by-step explanation:

    F can be rewritten as

    \bf F(x,y)=(6(x+y),6sin(y))

    The path C on which we are going to evaluate the integral is

    \bf (5,0)\overset{C_1}{\rightarrow}(5,7)\overset{C_2}{\rightarrow}(-3,7)\overset{C_3}{\rightarrow}(-3,0)\overset{C_4}{\rightarrow}(5,0)

    All the paths can be parametrized with parameter t (0≤ t≤ 1)

    Parametrization of \bf C_1

    r(t) = (5,0)+t((5,7)-(5,0)) = (5,0)+t(0,7) = (5,7t)

    \bf \displaystyle\int_{C_1}F=\displaystyle\int_{0}^{1}F(r(t))\bullet r'(t)dt=\displaystyle\int_{0}^{1}(6(5+7t ),6sin(7t))\bullet (0,7)dt=\\\\42\displaystyle\int_{0}^{1}sin(7t)dt\approx1.47672

    Parametrization of \bf C_2

    r(t) = (5,7)+t((-3,7)-(5,7)) = (5,7)+t(-8,0) = (5-8t,7)

    \bf \displaystyle\int_{C_2}F=\displaystyle\int_{0}^{1}F(r(t))\bullet r'(t)dt=\displaystyle\int_{0}^{1}(6(12-8t),6sin(7))\bullet (-8,0)dt=\\\\-48\displaystyle\int_{0}^{1}(12-8t)dt=-384

    Parametrization of \bf C_3

    r(t) = (-3,7)+t((-3,0)-(-3,7)) = (-3,7)+t(0,-7) = (-3,7-7t)

    \bf \displaystyle\int_{C_3}F=\displaystyle\int_{0}^{1}F(r(t))\bullet r'(t)dt=\displaystyle\int_{0}^{1}(6(4-7t),6sin(-7t))\bullet (0,-7)dt=\\\\-42\displaystyle\int_{0}^{1}sin(-7t)dt\approx1.47672

    Parametrization of \bf C_4

    r(t) = (-3,0)+t((5,0)-(-3,0)) = (-3,0)+t(8,0) = (-3+8t,0)

    \bf \displaystyle\int_{C_4}F=\displaystyle\int_{0}^{1}F(r(t))\bullet r'(t)dt=\displaystyle\int_{0}^{1}(6(-3+8t),6sin(0))\bullet (8,0)dt=\\\\48\displaystyle\int_{0}^{1}(-3+8t)dt=48

    Finally

    \bf \displaystyle\int_{C}F=\displaystyle\int_{C_1}F+\displaystyle\int_{C_2}F+\displaystyle\int_{C_3}F+\displaystyle\int_{C_4}F=\\\\\approx 1.47672-384+1.47672+48=-333.04656

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