# Evaluate the line integral oint_C xy dx + x^2 dy, 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.

Question
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.

2020-12-03
Step 1
Given,
$$\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}$$, Where C is the path going counterclockwise around the boundary of the rectangle with corners (0,0),(2,0),(2,3),(0,3)
Step 2
Green's theorem:
Let C be the simply connected region with piecewise smooth boundary C oriented counterclockwise,
If M and N are the first partial derivative in an open region containing C, then
$$\displaystyle\oint_{{C}}{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}=\int_{{R}}\int\frac{{\partial{N}}}{{\partial{x}}}-\frac{{\partial{M}}}{{\partial{y}}}{d}{A}$$
Consider, $$\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}$$
M=xy
$$\displaystyle{N}={x}^{{2}}$$
$$\displaystyle\frac{{\partial{M}}}{{\partial{y}}}={x}$$
$$\displaystyle\frac{{\partial{N}}}{{\partial{x}}}={2}{x}$$
Limits of x are $$\displaystyle{0}\le{x}\le{2}$$
Limits of y are $$\displaystyle{0}\le{y}\le{3}$$
Now the integral becomes,
$$\displaystyle{\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({2}{x}-{x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={\int_{{0}}^{{3}}}{\int_{{0}}^{{2}}}{\left({x}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}{{\left(\frac{{x}^{{2}}}{{2}}\right)}_{{0}}^{{2}}}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}{\left(\frac{{2}^{{2}}}{{2}}-\frac{{0}}{{2}}\right)}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}\frac{{1}}{{4}}{\left.{d}{y}\right.}$$
$$\displaystyle=\frac{{1}}{{4}}{\int_{{0}}^{{3}}}{\left.{d}{y}\right.}$$
$$\displaystyle=\frac{{1}}{{4}}{{\left({y}\right)}_{{0}}^{{3}}}$$
$$\displaystyle=\frac{{1}}{{4}}{\left({3}-{0}\right)}$$
$$\displaystyle=\frac{{3}}{{4}}$$
Therefore, $$\displaystyle\oint_{{C}}{x}{y}{\left.{d}{x}\right.}+{x}^{{2}}{\left.{d}{y}\right.}=\frac{{3}}{{4}}$$

### 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 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 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 $$\displaystyle\int_{{C}}{\left({e}^{{x}}+{y}^{{2}}\right)}{\left.{d}{x}\right.}+{\left({e}^{{y}}+{x}^{{2}}\right)}{\left.{d}{y}\right.}$$ where C is the boundary of the region(traversed counterclockwise) in the first quadrant bounded by $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={4}$$.
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.
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 $$\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}$$.
$$\displaystyle\oint_{{C}}{F}{8}{d}{r}$$, where $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{x}^{{2}},{x}^{{2}}\right\rangle}$$ and C consists of the arcs $$\displaystyle{y}={x}^{{2}}{\quad\text{and}\quad}{y}={8}{x}{f}{\quad\text{or}\quad}{0}\le{x}\le{8}$$
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).
$$\displaystyle\oint_{{C}}{\left({x}^{{2}}+{y}\right)}{\left.{d}{x}\right.}-{\left({3}{x}+{y}^{{3}}\right)}{\left.{d}{y}\right.}$$
Where c is the ellipse $$\displaystyle{x}^{{2}}+{4}{y}^{{2}}={4}$$
Use Green's Theorem to evaluate $$\displaystyle\int_{{C}}\vec{{{F}}}\cdot{d}\vec{{{r}}}$$ 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).