Answer:
Using Green's theorem we get,
[tex]I= \frac{-81\pi}{2}[/tex]
Step-by-step explanation:
Given function is F(x,y)=[tex](e^{2x} +x^{2}y,e^{2y} -y^{2}x)[/tex]
and path C is circle : [tex]x^{2} +y^{2} =9[/tex] clockwise.
Theory : Green's theorem is given by
[tex]I= \int\int\limits_C (\frac{\delta Q}{\delta x} -\frac{\delta P}{\delta y}) \, dx dy[/tex]
Where C is region bounded by curve in counterclockwise.
and F(x,y) = (P,Q)
Using Green's theorem,
F(x,y)=[tex](e^{2x} +x^{2}y,e^{2y} +y^{2}x)[/tex] and F(x,y) = (P,Q)
P=[tex]e^{2x} +x^{2}y[/tex]
Q=[tex]e^{2y} -y^{2}x[/tex]
Now,
[tex]\frac{\delta P}{\delta y}=\frac{\delta }{\delta y}(e^{2x} +x^{2}y)[/tex]
[tex]\frac{\delta P}{\delta y}=\frac{\delta }{\delta y}e^{2x} +\frac{\delta }{\delta y}x^{2}y[/tex]
[tex]\frac{\delta P}{\delta y}=0+x^{2}[/tex]
Also,
[tex]\frac{\delta Q}{\delta x}=\frac{\delta }{\delta x}(e^{2y} -y^{2}x)[/tex]
[tex]\frac{\delta Q}{\delta x}=\frac{\delta }{\delta x}e^{2y} -\frac{\delta }{\delta x}y^{2}x[/tex]
[tex]\frac{\delta Q}{\delta x}=-y^{2}[/tex]
Replacing the value in Green's theorem,
[tex]I= \int\int\limits_C (\frac{\delta Q}{\delta x} -\frac{\delta P}{\delta y}) \, dx dy[/tex]
[tex]I= \int\int\limits_C ( y^{2}+x^{2}) \, dx dy[/tex]
Introducing polar coordinates.
Take x=rcosu
y=rsinu
dxdy=rdrdu
Limits : r=0 to r=3
: u=0 to u={tex}2\pi{/tex}
[tex]I= \int\limits^{2\pi}_0\int\limits^3_0 ((rsinu)^{2}+(rcosu)^{2}) \, rdrdu\\I= \int\limits^{2\pi}_0\int\limits^3_0 [(r^{2}sin^{2}u)+(r^{2}cos^{2}u)] \, rdrdu\\I= \int\limits^{2\pi}_0\int\limits^3_0 [r^{2}] \, rdrdu\\I= \int\limits^{2\pi}_0\int\limits^3_0[r^{3}] \, drdu\\I= \int\limits^{2\pi}_0[\frac{r^{4}}{4}]^3_0 \, drdu\\I= \int\limits^{2\pi}_0[\frac{3^{4}}{4}] \, drdu\\I=\frac{3^{4}}{4}(2\pi-0)\\\\I= \frac{81}{4}(2\pi)[/tex]
Given curve C is clockwise.
Therefore,
[tex]I= (-1)\frac{81}{4}(2\pi)[/tex]
[tex]I= \frac{-81}{4}(2\pi)[/tex]
[tex]I= \frac{-81\pi}{2}[/tex]
