Solution :
Along the edge [tex]$C_1$[/tex]
The parametric equation for [tex]$C_1$[/tex] is given :
[tex]$x_1(t) = 9t , y_2(t) = 0 \ \ for \ \ 0 \leq t \leq 1$[/tex]
Along edge [tex]$C_2$[/tex]
The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain [tex]$0 \leq t \leq 1 $[/tex] is then given by :
[tex]$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$[/tex]
[tex]$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$[/tex]
Along edge [tex]$C_3$[/tex]
The parametric equation for [tex]$C_3$[/tex] is :
[tex]$x_1(t) = 0, \ \ \ y_2(t) = 9t \ \ \ for \ 0 \leq t \leq 1$[/tex]
Now,
x = 9t, ⇒ dx = 9 dt
y = 0, ⇒ dy = 0
[tex]$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$[/tex]
And
[tex]$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$[/tex]
[tex]$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$[/tex]
Then :
[tex]$\int_{C_1} y^2 x dx + x^2 y dy$[/tex]
[tex]$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$[/tex]
[tex]$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$[/tex]
= 0
And
x = 0, ⇒ dx = 0
y = 9 t, ⇒ dy = 9 dt
[tex]$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$[/tex]
Therefore,
[tex]$ \oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx $[/tex]
�� = 0 + 0 + 0
Applying the Green's theorem
[tex]$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$[/tex]
[tex]$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy $[/tex]
Here,
[tex]$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$[/tex]
[tex]$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$[/tex]
[tex]$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$[/tex]
Therefore,
[tex]$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$[/tex]
[tex]$= \int_0^9 0\ dy = 0$[/tex]
The vector field F is = [tex]$y^2 x \hat i+x^2 y \hat j$[/tex] is conservative.
