johnweldon1993
  • johnweldon1993
Verify Green Theorem Haven't worked with it in a while and having trouble with this one
Mathematics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
johnweldon1993
  • johnweldon1993
Verify Greens Theorem \[\large F = <6y^2 , 2x - 2y^4>\] Where R is the square with vertices (2,2),(2,-2),(-2,2),(-2,-2) So we know green theorem is \[\large \int_{C} F \cdot dr = \int \int_{R} [(F_2)_x - (F_1)_y] dA\] Let's do the RHS first \[\large \int \int_{R} [(F_2)_x - (F_1)_y] dxdy\] \[\large \int_{-2}^{2} \int_{-2}^{2} 2 - 12y dxdy\] \[\large \int_{-2}^{2} (2x - 12xy) |_{-2}^{2}) dy\] \[\large \int_{-2}^{2} 8 - (48y)dy\] \[\large 8y - 24y^2 |_{-2}^{2}\] \[\large 32 - 192 = -160\] Okay now the left hand side..the line integral \[\large \int_C F \cdot dr = \int_C(F_1dx + F_2dy)\] Break it into the 4 sides \[\large \int_{C1} (F_1 dx + F_2 dy) + \int_{C2} (F_1 dx + F_2 dy)+ \\ \int_{C3} (F_1 dx + F_2 dy)+ \int_{C4}(F_1 dx + F_2 dy)\] For C1 and C3 there is no change in 'x' so dx = 0 And for C2 and C4 dy = 0 So we have \[\large \int_{C1} F_2 dy + \int_{C2} F_1 dx+ \\ \int_{C3} F_2 dy+ \int_{C4}F_1 dx\] So \[\large \int_{C1} 2x - 2y^4 dy + \int_{C2} 6y^2 dx+ \\ \int_{C3} 2x - 2y^4 dy+ \int_{C4} 6y^2 dx\] C1 x = 2 C3 x = -2 C2 y = 2 C4 y = -2 \[\large \int_{C1} 4 - 2y^4 dy + \int_{C2} 24 dx+ \int_{C3} -4 - 2y^4 dy+ \int_{C4} 24 dx\] \[\large (4y - \frac{2}{5}y^5)|_{-2}^{2} + 24x|_{2}^{-2} + (-4y - \frac{2}{5}y^5)|_{2}^{-2} + 24x|_{-2}^{2}\] \[\large (16 - \frac{128}{5}) + (-48 - 48) + (16 + \frac{128}{5}) + (48 + 48)\] Cancelling out what you can I get \[\large 32\] Where am I going off? Does the fact that it encloses the origin change anything? What am I overlooking?
johnweldon1993
  • johnweldon1993
And lets just throw a diagram up with it quick |dw:1446671420652:dw|
johnweldon1993
  • johnweldon1993
You know what, its odd but if I change it so C4 is actually the x-axis...and multiply the result from that by 2...I get the same answer...so is that is? Since the square encloses the Origin....we work the side above the 'x-axis' and then multiply the result by 2?

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johnweldon1993
  • johnweldon1993
@phi Mind setting me straight quick?
phi
  • phi
you should be getting 32 for the right-hand side you did the integral correctly, so redo the evaluation of 8y - 24y^2
phi
  • phi
the 24y^2 term cancels out
johnweldon1993
  • johnweldon1993
\[\large (8y - 24y^2)|_{-2}^{2}\] \[\large (8(2) - 24(2)^2) - (8(-2) - 24(-2)^2)\] \[\large 16 - 96 + 16 + 96\] -_- Thank you @phi lol

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