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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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madhu.mukherjee.946
 one year ago
Best ResponseYou've already chosen the best response.1what is your question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Verify Green's Theorem for \[\oint_\limits C (x^22xy)dx+(x^2y+3)dy\] Where C is the boundary of \[y^2=8x\] and \[x=2\]dw:1442047557099:dw \[x=\frac{y^2}{8} \implies dx=\frac{y}{4}dy\] \[(x^22xy)dx+(x^2y+3)dy=(\frac{y^4}{4^3}\frac{y^3}{4})\frac{y}{4}dy+(\frac{y^5}{4^3}+3)dy\] \[\implies (x^22xy)dx+(x^2y+3)dy=(\frac{y^5}{4^4}\frac{y^4}{4^2}+\frac{y^5}{4^3}+3)dy\]\[\implies (x^22xy)dx+(x^2y+3)dy=(\frac{5y^5}{4^4}\frac{y^4}{4^2}+3)dy\] \[\oint \limits_C=\int\limits \limits_{C_{1}}+\int\limits \limits_{C_{2}}+\int\limits \limits_{C_{3}}\] 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 \[\therefore \int\limits_{C_{2}}=\int\limits_{4}^{4}(4y+3)dy=[2y^2+3y]_{4}^{4}=32+1232+12=24\] 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 \[\phi(x,y)=x^22xy, \psi(x,y)=x^2y+3\] \[\therefore \frac{\partial \psi}{\partial x}=2xy \space \space \space \space , \space \frac{\partial \phi}{\partial y}=2x\] 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 *_*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442050765061:dw Right? But for \(\sf C_1\) you have integrated from \(0\rightarrow 4\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh sorry, \(\sf 0\rightarrow 4\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442050859105:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, that's not my C1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I should've marked C1, C2, C3 earlier, I realized that later

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then my surface integral is wrong xD? At least I know where I am wrong

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i did it this way \[\int\limits_{x = 0}^{2} \ \int\limits_{y=\sqrt{8x}}^{\sqrt{8x}} 2xy + 2x \ dy \ dx \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then this means my limits are wrong for double integral....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442052380820:dw so blue then red.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or 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!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeh 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

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i've done it the other way, your way too let me now look at yours

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the integration limits should be \(\frac{y^2}{8} \lt x \lt 2\) and \(4 \lt y \lt 4\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhhhh OOPS!!!!! XD I can now finally rest in peace

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the power of Green !!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442053106423:dw ahh i get it now.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Jhannybean yep! nice drawing, that's how i do it too! saw it on a video on you tube  changed my life. **literally** lol!!!
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