# Use Green's Theorem to evaluate the line integral along the given positively oriented curve. int_C xy^2 dx + 4x^2y dy C is the triangle with vertices (0, 0), (3, 3), and (3, 6)

Question
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)

2020-12-04
Step 1
By Green's Theorem, we have
$$\displaystyle\int_{{C}}{P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}=\int_{{R}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{d}{A}$$
$$\displaystyle{P}{\left({x},{y}\right)}={x}{y}^{{2}}$$
$$\displaystyle{Q}{\left({x},{y}\right)}={4}{x}^{{2}}{y}$$
$$\displaystyle\frac{{\partial{Q}}}{{\partial{x}}}={8}{x}{y}$$
$$\displaystyle\frac{{\partial{P}}}{{\partial{y}}}={2}{x}{y}$$
$$\displaystyle\oint_{{C}}{\left({x}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}^{{2}}{y}{\left.{d}{y}\right.}\right)}=\int\int_{{R}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{d}{A}$$
$$\displaystyle=\int\int_{{R}}{\left({8}{x}{y}-{2}{x}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
Step 2
For line equation
$$\displaystyle{\left({y}-{3}\right)}=\frac{{{3}-{0}}}{{{3}-{0}}}{\left({x}-{3}\right)}\Rightarrow{y}={x}$$
$$\displaystyle{\left({y}-{6}\right)}=\frac{{{6}-{0}}}{{{3}-{0}}}{\left({x}-{3}\right)}\Rightarrow{y}={2}{x}$$
and x varies from 0 to 3
from i $$\displaystyle\oint_{{C}}{\left({x}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}^{{2}}{y}{\left.{d}{y}\right.}\right)}={\int_{{0}}^{{3}}}{\int_{{x}}^{{{2}{x}}}}{6}{x}{y}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}{\left[{6}{x}{{\mid}_{{x}}^{{{2}{x}}}}{y}{\left.{d}{y}\right.}\right]}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}{6}{x}{{\left[\frac{{y}^{{2}}}{{2}}\right]}_{{x}}^{{{2}{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}\frac{{{6}{x}}}{{2}}{\left[{4}{x}^{{2}}-{x}^{{2}}\right]}{\left.{d}{x}\right.}={\int_{{0}}^{{3}}}{9}{x}^{{3}}{\left.{d}{x}\right.}$$
$$\displaystyle={9}{\int_{{0}}^{{3}}}{x}^{{3}}{\left.{d}{x}\right.}={9}{{\left[\frac{{x}^{{4}}}{{4}}\right]}_{{0}}^{{3}}}=\frac{{{9}\cdot{81}}}{{4}}={182.25}$$

### 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
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).
(a)Use the Green's Theorem.
(b)Do not use the Green's Theorem.
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.
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 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.
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.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}{y}{i}+{3}{z}{j}+{5}{y}{k}$$, C is the curve of intersection of the plane x+z=10 and the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={9}$$.
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 Stokes' Theorem to compute $$\displaystyle\oint_{{C}}\frac{{1}}{{2}}{z}^{{2}}{\left.{d}{x}\right.}+{\left({x}{y}\right)}{\left.{d}{y}\right.}+{2020}{\left.{d}{z}\right.}$$, where C is the triangle with vertices at(1,0,0),(0,2,0), and (0,0,2) traversed in the order.
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 $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}$$.