Answer:
[tex]\mathbf{- 20 \pi + \dfrac{40}{3}}[/tex]
Step-by-step explanation:
Given that:
[tex]\int\limits_c {F} \, dr[/tex]
where;
[tex]F(x,y) = \langle\sqrt{x} + 10y^3+10y^2 + \sqrt{y} \rangle[/tex] 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.
[tex]\int \limits_CF. dr = \iint_D \Biggl \langle \dfrac{\partial Q}{\partial X} - \dfrac{\partial P }{\partial Y} \Biggl \rangle dA[/tex]
Then;
[tex]F(P,Q) = \langle\sqrt{x} + 10y^3+10y^2 + \sqrt{y} \rangle[/tex]
[tex]\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[/tex]
[tex]\int \limits _CF. dr = -\iint_D \Biggl \langle 10(2x)-10(3y^2) \Biggl \rangle dA[/tex]
[tex]\int \limits _CF. dr = \iint_D \Big \langle -20x+30y^2 \Big \rangle dA[/tex]
y → 0 to sin (x) x → 0 to π
[tex]\int \limits _CF. dr = \int \limits ^{\pi}_{0} \int \limits ^{sin \ x }_{0} \Big \langle -20x+30y^2 \Big \rangle dydx[/tex]
[tex]\int \limits _CF. dr = \int \limits ^{\pi}_{0} \Biggl [ -20xy + \dfrac{30 \ y^3}{3} \Biggl ] ^{sin\ x}_{0} \ dx[/tex]
[tex]\int \limits _CF. dr = \int \limits ^{\pi}_{0} \Big [ -20x \ sin x + 10 \ sin^3x \Big ] \ dx[/tex]
replace [tex]sin^3 x = \dfrac{3}{4} \ sin x - \dfrac{1}{4} \ sin (3x)[/tex]
[tex]= \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[/tex]
By applying integration by posits
[tex]\int (u)(v') \ dx = (u)(v) - \int (u') (v) \ dx \\ \\ u = -20x \ \ \ \ \ \ \ v'= sin \ x \\ \\ u' = -20 \ \ \ \ \ \ \ v = - cos \ x[/tex]
[tex]= (-20x) (-cos x) - \int (-20)(-cos x) \ dx + 10 \Big [\dfrac{3}{4} \ cos x + \dfrac{1}{12}\ cos (3x) \Big ][/tex]
[tex]= 20x \ cos x - 20 \ sin x- \dfrac{15}{2} \ cos \ x + \dfrac{5}{6} \ cos (3x) \Biggl |^{\pi}_{0}[/tex]
[tex]= ( -20 \pi -0 +\dfrac{15}{2}-\dfrac{5}{6}) - (0-0-\dfrac{15}{2}+\dfrac{5}{6})[/tex]
[tex]= - 20 \pi + \dfrac{15}{2}-\dfrac{5}{6}+\dfrac{15}{2}-\dfrac{5}{6}[/tex]
[tex]\mathbf{= - 20 \pi + \dfrac{40}{3}}[/tex]
