Question

Use Green's Theorem to evaluate C F · dr. (Check the orientation of the curve before applying the theorem.) F(x, y) = x + 10y3, 10x2 + y C consists of the arc of the curve y = sin(x) from (0, 0) to (π, 0) and the line segment from (π, 0) to (0, 0)

Answer 1

Answer:

$\mathbf{- 20 \pi + \dfrac{40}{3}}$

Step-by-step explanation:

Given that:

$\int\limits_c {F} \, dr$

where;

$F(x,y) = \langle\sqrt{x} + 10y^3+10y^2 + \sqrt{y} \rangle$ and C consist of the arc of the curve, This shows that C is a closed curve.

Thus, using Green's theorem for clockwise orientation.

$\int \limits_CF. dr = \iint_D \Biggl \langle \dfrac{\partial Q}{\partial X} - \dfrac{\partial P }{\partial Y} \Biggl \rangle dA$

Then;

$F(P,Q) = \langle\sqrt{x} + 10y^3+10y^2 + \sqrt{y} \rangle$

$\int \limits _CF. dr = -\iint_D \Biggl \langle \dfrac{\partial}{\partial} ( 10x^2 + \sqrt{y} -\dfrac{\partial}{\partial y } ( \sqrt{x} + 10y^3) \Biggl \rangle dA$

$\int \limits _CF. dr = -\iint_D \Biggl \langle 10(2x)-10(3y^2) \Biggl \rangle dA$

$\int \limits _CF. dr = \iint_D \Big \langle -20x+30y^2 \Big \rangle dA$

y → 0 to sin (x)        x → 0 to π

$\int \limits _CF. dr = \int \limits ^{\pi}_{0} \int \limits ^{sin \ x }_{0} \Big \langle -20x+30y^2 \Big \rangle dydx$

$\int \limits _CF. dr = \int \limits ^{\pi}_{0} \Biggl [ -20xy + \dfrac{30 \ y^3}{3} \Biggl ] ^{sin\ x}_{0} \ dx$

$\int \limits _CF. dr = \int \limits ^{\pi}_{0} \Big [ -20x \ sin x + 10 \ sin^3x \Big ] \ dx$

replace $sin^3 x = \dfrac{3}{4} \ sin x - \dfrac{1}{4} \ sin (3x)$

$= \int \limits ^{\pi}_{0} -20 x sin x dx + \int \limits ^{\pi}_{0} 10(\dfrac{3}{4} sin x - \dfrac{1}{4} sin (3x) ) \ dx$

By applying integration by posits

$\int (u)(v') \ dx = (u)(v) - \int (u') (v) \ dx \\ \\ u = -20x \ \ \ \ \ \ \ v'= sin \ x \\ \\ u' = -20 \ \ \ \ \ \ \ v = - cos \ x$

$= (-20x) (-cos x) - \int (-20)(-cos x) \ dx + 10 \Big [\dfrac{3}{4} \ cos x + \dfrac{1}{12}\ cos (3x) \Big ]$

$= 20x \ cos x - 20 \ sin x- \dfrac{15}{2} \ cos \ x + \dfrac{5}{6} \ cos (3x) \Biggl |^{\pi}_{0}$

$= ( -20 \pi -0 +\dfrac{15}{2}-\dfrac{5}{6}) - (0-0-\dfrac{15}{2}+\dfrac{5}{6})$

$= - 20 \pi + \dfrac{15}{2}-\dfrac{5}{6}+\dfrac{15}{2}-\dfrac{5}{6}$

$\mathbf{= - 20 \pi + \dfrac{40}{3}}$