Evaluate int_C x^2y^2 dx + 4xy^3 dy 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 int_C x^2y^2 dx + 4xy^3 dy 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.

Question
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.

Answers (1)

2021-02-26
Step 1
Given:
int_c x^2y^2dx+4xy^3dy
where c is the triangle with vertices (0,0),(1,3),(0,3)
Step 2
\(\displaystyle\int_{{c}}{x}^{{2}}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}{y}^{{3}}{\left.{d}{y}\right.}\)
co comparing with \(\displaystyle\int_{{c}}{P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}\) we have
\(\displaystyle{P}={x}^{{2}}{y}^{{2}}{\quad\text{and}\quad}{Q}={4}{x}{y}^{{3}}\)
By Green's Theorem
\(\displaystyle\int_{{c}}{x}^{{2}}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}{y}^{{3}}{\left.{d}{y}\right.}=\int\int_{{D}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{d}{A}\)
\(\displaystyle=\int\int_{{D}}{\left({4}{y}^{{3}}-{2}{x}^{{2}}{y}\right)}{d}{A}\)
\(\displaystyle{D}={\left[{\left({x},{y}\right)}:{0}\le{x}\le{1},{3}{x}\le{y}\le{3}\right]}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\int_{{{3}{x}}}^{{3}}}{\left({4}{y}^{{3}}-{2}{x}^{{2}}{y}\right)}{\left.{d}{y}\right.}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{{\left({y}^{{4}}-{x}^{{2}}{y}^{{1}}\right)}_{{{y}={3}{x}}}^{{{y}={3}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\left({81}-{9}{x}^{{2}}-{72}{x}^{{4}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left({81}{x}-{3}{x}^{{3}}-\frac{{72}}{{5}}{x}^{{5}}\right)}_{{0}}^{{1}}}\)
\(\displaystyle=\frac{{318}}{{5}}\)
Step 3
\(\displaystyle\int_{{c}}{x}^{{2}}{y}^{{2}}{\left.{d}{x}\right.}+{4}{x}{y}^{{3}}{\left.{d}{y}\right.}\)
\(\displaystyle{c}_{{1}}:{y}={3}{x}\)
dy=3dx
\(\displaystyle{0}\le{x}\le{1}\)
\(\displaystyle{c}_{{2}}:{y}={3}\)
dy = 0
x from 1 to 0
\(\displaystyle{c}_{{3}}:{x}={0}\)
dx = 0
y from 3 to 0
Hence
\(\displaystyle{I}={\int_{{0}}^{{1}}}{x}^{{2}}{\left({3}{x}\right)}^{{2}}{\left.{d}{x}\right.}+{4}{x}{\left({3}{x}\right)}^{{3}}{\left.{d}{x}\right.}+{\int_{{1}}^{{0}}}{9}{x}^{{2}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\left({9}{x}^{{4}}+{324}{x}^{{4}}\right)}{\left.{d}{x}\right.}+{{\left[\frac{{{9}{x}^{{3}}}}{{3}}\right]}_{{1}}^{{0}}}\)
\(\displaystyle={{\left[\frac{{{333}{x}^{{5}}}}{{5}}\right]}_{{0}}^{{1}}}-{3}\)
\(\displaystyle=\frac{{318}}{{5}}\)
0

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