anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
madhu.mukherjee.946
  • madhu.mukherjee.946
what is your question
anonymous
  • 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 *_*
anonymous
  • anonymous
@IrishBoy123

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Jhannybean
  • Jhannybean
*
Jhannybean
  • Jhannybean
|dw:1442050765061:dw| Right? But for \(\sf C_1\) you have integrated from \(0\rightarrow 4\)?
Jhannybean
  • Jhannybean
Oh sorry, \(\sf 0\rightarrow -4\)?
anonymous
  • anonymous
|dw:1442050859105:dw|
anonymous
  • anonymous
no, that's not my C1
Jhannybean
  • Jhannybean
Ok, I see
anonymous
  • anonymous
lol :p
anonymous
  • anonymous
I should've marked C1, C2, C3 earlier, I realized that later
Jhannybean
  • Jhannybean
Yeah xD
IrishBoy123
  • 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
anonymous
  • anonymous
Then my surface integral is wrong xD? At least I know where I am wrong
IrishBoy123
  • IrishBoy123
i did it this way \[\int\limits_{x = 0}^{2} \ \int\limits_{y=-\sqrt{8x}}^{\sqrt{8x}} 2xy + 2x \ dy \ dx \]
anonymous
  • anonymous
Then this means my limits are wrong for double integral....
Jhannybean
  • Jhannybean
|dw:1442052380820:dw| so blue then red.
Jhannybean
  • 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!!!!
anonymous
  • 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
IrishBoy123
  • IrishBoy123
i've done it the other way, your way too let me now look at yours
IrishBoy123
  • IrishBoy123
the integration limits should be \(\frac{y^2}{8} \lt x \lt 2\) and \(-4 \lt y \lt 4\)
anonymous
  • anonymous
ohhhhh OOPS!!!!! XD I can now finally rest in peace
IrishBoy123
  • IrishBoy123
the power of Green !!!!
Jhannybean
  • Jhannybean
|dw:1442053106423:dw| ahh i get it now.
IrishBoy123
  • 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!!!

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