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 =_________.

Answer 1

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

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