## anonymous one year ago ques

2. anonymous

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 $x=2$|dw:1442047557099:dw| $x=\frac{y^2}{8} \implies dx=\frac{y}{4}dy$ $(x^2-2xy)dx+(x^2y+3)dy=(\frac{y^4}{4^3}-\frac{y^3}{4})\frac{y}{4}dy+(\frac{y^5}{4^3}+3)dy$ $\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 *_*

3. anonymous

@IrishBoy123

4. Jhannybean

*

5. Jhannybean

|dw:1442050765061:dw| Right? But for $$\sf C_1$$ you have integrated from $$0\rightarrow 4$$?

6. Jhannybean

Oh sorry, $$\sf 0\rightarrow -4$$?

7. anonymous

|dw:1442050859105:dw|

8. anonymous

no, that's not my C1

9. Jhannybean

Ok, I see

10. anonymous

lol :p

11. anonymous

I should've marked C1, C2, C3 earlier, I realized that later

12. Jhannybean

Yeah xD

13. IrishBoy123

i get 128/5 for the area integral haven't done the line integral) but i did it dy dx so let's look at that first

14. anonymous

Then my surface integral is wrong xD? At least I know where I am wrong

15. IrishBoy123

i did it this way $\int\limits_{x = 0}^{2} \ \int\limits_{y=-\sqrt{8x}}^{\sqrt{8x}} 2xy + 2x \ dy \ dx$

16. anonymous

Then this means my limits are wrong for double integral....

17. Jhannybean

|dw:1442052380820:dw| so blue then red.

18. Jhannybean

or no, backwards! red then blue. whoops. Red represents $$\sf y=\pm \sqrt{8x}$$ and then blue integrates horizontally $$\sf 0\rightarrow 3$$ . That makes it SOOOO much simpler!!!!

19. anonymous

Yeh but, $\iint_\limits{R}f(x,y)dxdy=\iint_\limits{R}f(x,y)dydx$ So I should get the same answer either way? This can only mean my limits are wrong

20. IrishBoy123

i've done it the other way, your way too let me now look at yours

21. IrishBoy123

the integration limits should be $$\frac{y^2}{8} \lt x \lt 2$$ and $$-4 \lt y \lt 4$$

22. anonymous

ohhhhh OOPS!!!!! XD I can now finally rest in peace

23. IrishBoy123

the power of Green !!!!

24. Jhannybean

|dw:1442053106423:dw| ahh i get it now.

25. IrishBoy123

@Jhannybean yep! nice drawing, that's how i do it too! saw it on a video on you tube -- changed my life. **literally** lol!!!