Question

ques

Answer 1

what is your question

Answer 2

Verify Green's Theorem for
\[\oint_\limits C (x^2-2xy)dx+(x^2y+3)dy\]
Where C is the boundary of \[y^2=8x\]
and
\[x=2\]
\[x=\frac{y^2}{8} \implies dx=\frac{y}{4}dy\]
\[(x^2-2xy)dx+(x^2y+3)dy=(\frac{y^4}{4^3}-\frac{y^3}{4})\frac{y}{4}dy+(\frac{y^5}{4^3}+3)dy\]
\[\implies (x^2-2xy)dx+(x^2y+3)dy=(\frac{y^5}{4^4}-\frac{y^4}{4^2}+\frac{y^5}{4^3}+3)dy\]\[\implies (x^2-2xy)dx+(x^2y+3)dy=(\frac{5y^5}{4^4}-\frac{y^4}{4^2}+3)dy\]
\[\oint \limits_C=\int\limits \limits_{C_{1}}+\int\limits \limits_{C_{2}}+\int\limits \limits_{C_{3}}\]
For C1 we have
\[\int\limits \limits_{C_{1}}=\int\limits_{0}^{-4}(\frac{5y^5}{4^4}-\frac{y^4}{4^2}+3)dy=[\frac{5y^6}{6\times4^4}-\frac{y^5}{5\times4^2}+3y]_{0}^{-4}\]\[\int\limits \limits_{C_{1}}=\frac{5}{6}.\frac{4^6}{4^4}+\frac{1}{5}.\frac{4^5}{4^2}-12=\frac{40}{3}+\frac{64}{5}-12\]
For C2 we have
x=2, dx=0
\[\therefore \int\limits_{C_{2}}=\int\limits_{-4}^{4}(4y+3)dy=[2y^2+3y]_{-4}^{4}=32+12-32+12=24\]
for C3 we have
\[\int\limits \limits_{C_{3}}=\int\limits_{4}^0(\frac{5y^5}{4^4}-\frac{y^4}{4^2}+3)dy=[\frac{5y^6}{6\times4^4}-\frac{y^5}{5\times4^2}+3y]_{4}^{0}\]
\[\int\limits \limits_{C_{3}}=-(\frac{5}{6}.\frac{4^6}{4^4}-\frac{1}{5}.\frac{4^5}{4^2}+12)=-\frac{40}{3}+\frac{64}{5}-12\]
\[\therefore \oint \limits_{C}=\frac{40}{3}+\frac{64}{5}-12+24-\frac{40}{3}+\frac{64}{5}-12=\frac{128}{5}\]
Let
\[\phi(x,y)=x^2-2xy, \psi(x,y)=x^2y+3\]
\[\therefore \frac{\partial \psi}{\partial x}=2xy \space \space \space \space , \space \frac{\partial \phi}{\partial y}=-2x\]
Applying Green's Theorem,
\[\oint_\limits{C}\phi dx+\psi dy=\iint_\limits{R}(\frac{\partial \psi}{\partial x}-\frac{\partial \phi}{\partial y})dxdy\]\[\oint_\limits{C}=\iint_\limits{R}(2xy+2x)dxdy=\int\limits_{-4}^{4}\int\limits_{0}^{\frac{y^2}{8}}(2xy+2x)dxdy=\frac{32}{5}\]
Where am I going wrong *_*