Question

# 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)

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 $$\displaystyle\oint_{{C}}{\left({x}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}^{{2}}{y}{\left.{d}{y}\right.}\right)}={\int_{{0}}^{{3}}}{\int_{{x}}^{{{2}{x}}}}{6}{x}{y}{\left.{d}{x}\right.}{\left.{d}{y}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}{\left[{6}{x}{{\mid}_{{x}}^{{{2}{x}}}}{y}{\left.{d}{y}\right.}\right]}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}{6}{x}{{\left[\frac{{y}^{{2}}}{{2}}\right]}_{{x}}^{{{2}{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{0}}^{{3}}}\frac{{{6}{x}}}{{2}}{\left[{4}{x}^{{2}}-{x}^{{2}}\right]}{\left.{d}{x}\right.}={\int_{{0}}^{{3}}}{9}{x}^{{3}}{\left.{d}{x}\right.}$$
$$\displaystyle={9}{\int_{{0}}^{{3}}}{x}^{{3}}{\left.{d}{x}\right.}={9}{{\left[\frac{{x}^{{4}}}{{4}}\right]}_{{0}}^{{3}}}=\frac{{{9}\cdot{81}}}{{4}}={182.25}$$