Question

Verify Green Theorem

Haven't worked with it in a while and having trouble with this one

Answer 1

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?

Answer 2

And lets just throw a diagram up with it quick

Answer 3

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?

Answer 4

@phi Mind setting me straight quick?

Answer 5

you should be getting 32 for the right-hand side
you did the integral correctly, so redo the evaluation of
8y - 24y^2

Answer 6

the 24y^2 term cancels out

Answer 7

\[\large (8y - 24y^2)|_{-2}^{2}\]

\[\large (8(2) - 24(2)^2) - (8(-2) - 24(-2)^2)\]
\[\large 16 - 96 + 16 + 96\]

-_- Thank you @phi lol