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.

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
${\displaystyle {\int }_c{x}^2{y}^{2}dx+4x{y}^{3}dy}$
co comparing with ${\displaystyle {\int }_{c}Pdx+Qdy}$ we have
${\displaystyle P=x^2{y}^2\text{and}Q=4x{y}^3}$
By Green's Theorem
${\displaystyle {\int }_c{x}^2{y}^{2}dx+4x{y}^{3}dy=\int {\int }_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA}$
${\displaystyle =\int {\int }_D(4y^3-2x^{2}y)dA}$
${\displaystyle D=[(x,y):0\le x\le 1,3x\le y\le 3]}$
${\displaystyle ={\int }_0^1{\int }_{3x}^3(4y^3-2x^{2}y)dydx}$
${\displaystyle ={\int }_0^1{(y^4-x^2{y}^1)}_{y=3x}^{y=3}dx}$
${\displaystyle ={\int }_0^1(81-9x^2-72x^4)dx}$
${\displaystyle ={(81x-3x^3-\frac{72}{5}{x}^5)}_0^1}$
${\displaystyle =\frac{318}{5}}$
Step 3
${\displaystyle {\int }_c{x}^2{y}^{2}dx+4x{y}^{3}dy}$
${\displaystyle c_1:y=3x}$
$dy=3dx$
${\displaystyle 0\le x\le 1}$
${\displaystyle c_2:y=3}$
$dy=0$
x from 1 to 0
${\displaystyle c_3:x=0}$
$dx=0$
y from 3 to 0
Hence
${\displaystyle I={\int }_0^1{x}^2{\left(3x\right)}^{2}dx+4x{}^{}}$