# Use Green's Theorem to evaluate F * dr. (Check the orientation of the curve before applying the theorem.) F(x, y) = sqrtx+ 4y^3, 4x^2 + sqrt(y) C consists of the arc of the curve y = sin(x) from (0, 0) to (pi, 0) and the line segment from (pi, 0) to (0, 0)

Question
Use Green's Theorem to evaluate F * dr. (Check the orientation of the curve before applying the theorem.)
$$\displaystyle{F}{\left({x},{y}\right)}=\sqrt{{x}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}$$
C consists of the arc of the curve $$\displaystyle{y}={\sin{{\left({x}\right)}}}$$ from (0, 0) to $$\displaystyle{\left(\pi,{0}\right)}$$ and the line segment from $$\displaystyle{\left(\pi,{0}\right)}$$ to (0, 0)

2021-01-14
Step 1
The given function is,
$$\displaystyle{F}{\left({x},{y}\right)}={\left(\sqrt{{{x}}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}\right)}$$
Step 2
C is a closed curve and using Green’s theorem for clockwise orientation the integral is evaluated using the below formula.
$$\displaystyle\oint_{{C}}{F}\cdot{d}{r}=-\int\int_{{D}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{d}{A}$$
Where, $$\displaystyle{F}={\left({P},{Q}\right)}={\left(\sqrt{{{x}}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}\right)}$$
Step 3
Calculate the partial derivative of P and Q as follows.
$$\displaystyle\frac{{\partial{Q}}}{{\partial{x}}}=\frac{{\partial}}{{\partial{x}}}{\left({4}{x}^{{2}}+\sqrt{{{y}}}\right)}$$
=8x
$$\displaystyle\frac{{\partial{P}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left(\sqrt{{{x}}}+{4}{y}^{{3}}\right)}$$
$$\displaystyle={12}{y}^{{2}}$$
Step 4
Evaluate the integral Fdr by substituting the limits as follows.
$$\displaystyle\oint_{{C}}{F}\cdot{d}{r}={\int_{{0}}^{\pi}}{\int_{{0}}^{{{\sin{{x}}}}}}{\left(-{8}{x}+{12}{y}^{{2}}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{\pi}}{{\left(-{8}{x}{y}+\frac{{{12}{y}^{{3}}}}{{3}}\right)}_{{0}}^{{{\sin{{x}}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{\pi}}{\left(-{8}{x}{\left({\sin{{x}}}\right)}+{4}{{\sin}^{{3}}{x}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle=-{\int_{{0}}^{\pi}}{8}{x}{\sin{{x}}}{\left.{d}{x}\right.}+{\int_{{0}}^{\pi}}{4}{{\sin}^{{3}}{x}}{\left.{d}{x}\right.}$$
$$\displaystyle=-{8}{{\left({x}{\left(-{\cos{{x}}}\right)}-\int{\cos{{x}}}{\left.{d}{x}\right.}\right)}_{{0}}^{\pi}}+{4}{\int_{{0}}^{\pi}}{{\sin}^{{2}}{x}}{\sin{{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle=-{8}{{\left(-{x}{\cos{{x}}}-{\sin{{x}}}\right)}_{{0}}^{\pi}}+{4}{\int_{{0}}^{\pi}}{\left({1}-{{\cos}^{{2}}{x}}\right)}{\sin{{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle=-{8}\pi+{4}{{\left(-{\cos{{x}}}-\frac{{{{\cos}^{{3}}{x}}}}{{3}}\right)}_{{0}}^{\pi}}$$
$$\displaystyle=-{8}\pi+{4}{\left({2}-\frac{{2}}{{3}}\right)}$$
$$\displaystyle=-{8}\pi+\frac{{16}}{{3}}$$

### Relevant Questions

Use Green's Theorem to evaluate the line integral. Orient the curve counerclockwise.
$$\displaystyle\oint_{{C}}{F}{8}{d}{r}$$, where $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{x}^{{2}},{x}^{{2}}\right\rangle}$$ and C consists of the arcs $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={8}{x}{f}{\quad\text{or}\quad}{0}\le{x}\le{8}$$
Let C be the ellipse contained in the xy plane whose equation is $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore W = 0.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that W = 12.56.
c) We can apply Stokes' Theorem and conclude that W = 6.28
d) We can apply Stokes' Theorem and conclude that W = 12.56.
Use Green’s Theorem to evaluate around the boundary curve C of the region R, where R is the triangle formed by the point (0, 0), (1, 1) and (1, 3).
Find the work done by the force field F(x,y)=4yi+2xj in moving a particle along a circle $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ from(0,1)to(1,0).
Use Green's Theorem to evaluate the line integral along the given positively oriented curve.
$$\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}^{{2}}{y}{\left.{d}{y}\right.}$$
C is the triangle with vertices (0, 0), (3, 3), and (3, 6)
Use Green's Theorem to evaluate $$\displaystyle\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$ where $$\displaystyle\vec{{{F}}}{\left({x},{y}\right)}={x}{y}^{{2}}{i}+{\left({1}-{x}{y}^{{3}}\right)}{j}$$ and C is the parallelogram with vertices (-1,2), (-1,-1),(1,1)and(1,4).
The orientation of C is counterclockwise.
Use Stokes' theorem to evaluate the line integral $$\displaystyle\oint_{{C}}{F}\cdot{d}{r}$$ where A = -yi + xj and C is the boundary of the ellipse $$\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}$$.
Use Stokes' Theorem to evaluate $$\displaystyle\int_{{C}}{F}\cdot{d}{r}$$ where C is oriented counterclockwise as viewed from above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}+{y}^{{2}}\right)}{i}+{\left({y}+{z}^{{2}}\right)}{j}+{\left({z}+{x}^{{2}}\right)}{k}$$,
C is the triangle with vertices (3,0,0),(0,3,0), and (0,0,3).
Use Stokes' Theorem to evaluate int_C F*dr where C is oriented counterclockwise as viewed above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}$$, C is the curve of intersection of the plane x+z=10 and the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={9}$$.
$$\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}$$
Let $$\displaystyle{f}={\left[{x}^{{2}}{y}^{{2}},-\frac{{x}}{{y}^{{2}}}\right]}$$ and $$\displaystyle{R}:{1}\le{x}^{{2}}+{y}^{{2}},+{4},{x}\ge{0},{y}\ge{x}$$. Evaluate $$\displaystyle\int_{{C}}{F}{\left({r}\right)}\cdot{d}{r}$$ counterclockwise around the boundary C of the region R by Green's theorem.