# 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 ${\int }_{C}{x}^{2}{y}^{2}dx+4x{y}^{3}dy$ where C is the triangle with vertices(0,0),(1,3), and (0,3).
(a)Use the Greens
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Step 1
Given:
${\int }_{c}{x}^{2}{y}^{2}dx+4x{y}^{3}dy$
where c is the triangle with vertices (0,0),(1,3),(0,3)
Step 2
${\int }_{c}{x}^{2}{y}^{2}dx+4x{y}^{3}dy$
co comparing with ${\int }_{c}Pdx+Qdy$ we have
$P={x}^{2}{y}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Q=4x{y}^{3}$
By Green's Theorem
${\int }_{c}{x}^{2}{y}^{2}dx+4x{y}^{3}dy=\int {\int }_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$
$=\int {\int }_{D}\left(4{y}^{3}-2{x}^{2}y\right)dA$
$D=\left[\left(x,y\right):0\le x\le 1,3x\le y\le 3\right]$
$={\int }_{0}^{1}{\int }_{3x}^{3}\left(4{y}^{3}-2{x}^{2}y\right)dydx$
$={\int }_{0}^{1}{\left({y}^{4}-{x}^{2}{y}^{1}\right)}_{y=3x}^{y=3}dx$
$={\int }_{0}^{1}\left(81-9{x}^{2}-72{x}^{4}\right)dx$
$={\left(81x-3{x}^{3}-\frac{72}{5}{x}^{5}\right)}_{0}^{1}$
$=\frac{318}{5}$
Step 3
${\int }_{c}{x}^{2}{y}^{2}dx+4x{y}^{3}dy$
${c}_{1}:y=3x$
$dy=3dx$
$0\le x\le 1$
${c}_{2}:y=3$
$dy=0$
x from 1 to 0
${c}_{3}:x=0$
$dx=0$
y from 3 to 0
Hence
$I={\int }_{0}^{1}{x}^{2}{\left(3x\right)}^{2}dx+4x{}^{}$