# Use Green's Theorem to evaluate the line integral. (2x + ln 2)dy - (8y^2 + sinhx)dx Question
Green's, Stokes', and the divergence theorem Use Green's Theorem to evaluate the line integral.
$$\displaystyle{\left({2}{x}+{\ln{{2}}}\right)}{\left.{d}{y}\right.}-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}{\left.{d}{x}\right.}$$ 2020-12-14
Step 1
Let's compare the given integrand with the standard form.
$$\displaystyle{\left({2}{x}+{\ln{{2}}}\right)}{\left.{d}{y}\right.}-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}{\left.{d}{x}\right.}={Q}.{\left.{d}{y}\right.}+{P}.{\left.{d}{x}\right.}$$
Hence, $$\displaystyle{Q}={2}{x}+{\ln{{2}}}$$
and $$\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}$$
Step 2
$$\displaystyle{P}=-{\left({8}{y}^{{2}}+{\text{sinh}{{x}}}\right)}$$
$$\displaystyle\Rightarrow\frac{{\partial{P}}}{{\partial{y}}}=-{16}{y}$$
$$\displaystyle{Q}={\left({2}{x}+{\ln{{2}}}\right)}$$
$$\displaystyle\Rightarrow\frac{{\partial{Q}}}{{\partial{x}}}={2}$$
Step 3
Limits of integration for x will be from x = 3 to x = 6 (the x corrdinates of the square)
and limits of integration for y will be from y = 1 to y = 4 (the y coordinates of the square)
$$\displaystyle{I}=\oint_{{C}}{\left({P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}\right)}=\int\int_{{R}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{d}{A}$$
$$\displaystyle=\int\int_{{R}}{\left({2}+{16}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{1}}^{{4}}}{\left({2}+{16}{y}\right)}{\left.{d}{y}\right.}{\int_{{3}}^{{6}}}{\left.{d}{x}\right.}$$
$$\displaystyle={{\left[{2}{y}+{8}{y}^{{2}}\right]}_{{1}}^{{4}}}\times{{\left[{x}\right]}_{{3}}^{{6}}}$$
$$\displaystyle={126}\times{3}={378}$$
Step 4
Given integral = 378

### Relevant Questions Use Green's Theorem to evaluate the line integral
$$\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}$$
where C is triangle with vertices (0,0),(0,2)and(2,2) oriented counterclockwise.
a)6
b)10
c)14
d)4
e)8
f)12 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) Evaluate the line integral $$\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}$$, where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3), and (0,3). You can evaluate directly or use Green's theorem.
Write the integral(s), but do not evaluate. Evaluate the line integral by the two following methods. y) dx + (x+y)dy C os counerclockwise around the circle with center the origin and radius 3(a) directly (b) using Green's Theorem. Use Green's Theorem to evaluate $$\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}$$ where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}$$. Use Green's Theorem to evaluate
$$\displaystyle\oint_{{C}}{\left({x}^{{2}}+{y}\right)}{\left.{d}{x}\right.}-{\left({3}{x}+{y}^{{3}}\right)}{\left.{d}{y}\right.}$$
Where c is the ellipse $$\displaystyle{x}^{{2}}+{4}{y}^{{2}}={4}$$ 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 $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={8}{x}{f}{\quad\text{or}\quad}{0}\le{x}\le{8}$$ Evaluate $$\displaystyle\int_{{C}}{x}^{{2}}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}{y}^{{3}}{\left.{d}{y}\right.}$$ where C is the triangle with vertices(0,0),(1,3), and (0,3). $$\displaystyle{F}{\left({x},{y}\right)}=\sqrt{{x}}+{4}{y}^{{3}},{4}{x}^{{2}}+\sqrt{{{y}}}$$
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) $$\displaystyle\oint_{{c}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$