Question

can someone verify green's theorem given the vector field F= (-x^2y)i + (xy^2)j and region D: x^2+y^2 =or< 4 (disk) using: integral Mdx +Ndy

Answer 1

\[\int\limits_{?}^{?}Mdx+Ndy\] for a cleaer representation

Answer 2

First doing it the long way.

Parametrise the curve

its a circle , so use polar coordinates


x= 2cos(t) , y=2sin(t) 0<=t<=2pi

Answer 3

therefore we can express the curve as r(t) , r(t) = (2cos(t)) i + (2sin(t))j = ( 2cos(t) , 2sin(t) )

Now , we want integral F dr

so we need the differential, dr ,

dr = ( -2sin(t) , 2cos(t) ) dt

Answer 4

now sub the parametrisation into our vector field F ( replace x with 2cos(t), and y with 2sin(t) )

so that gives \[F = ( -8 \cos^2(t)\sin(t) , 8\cos(t)\sin^2(t) ) \]

where I have used the bracket notation for the vector function.

Answer 5

now get F and dot that with dr \[F . dr = 16\cos^2(t)\sin^2(t) + 16\cos^2(t)\sin^2(t) = 32 \cos^2(t)\sin^2(t)\]

Answer 6

so thing you want(derived from first principles ) is\[32 \int\limits_{0}^{2 \pi } \sin^2(t)\cos^2(t) dt \]

Answer 7

you can do the integra, all you need is to used double angle formula for cos

Answer 8

now by greens theorm it says we can get the same answer by doing a double integral over the region \[=\int\limits \int\limits_{R} ( \frac{dQ}{dx} - \frac{dP}{dy} ) dA\]

where P is the first component of the vector field , and Q is the second.

Answer 9

\[\frac{dQ}{dx} - \frac{dP}{dy} = y^2-(-x^2) = y^2 +x^2\]

so we need to find the integral \[\int\limits \int\limits_{R} ( x^2 +y^2) dA\]

Answer 10

now, the domain is a disc, we change to polar coordinates

so x= rcos(t) , y= rsin(t) and the jacobian of the coordinate change is r.

The area in polar co ordinates is \[0 \le r \le 2 ...... 0 \le \theta \le 2 \pi\]

Answer 11

so the answer is also equal to \[\int\limits_{0}^{2\pi} d \theta \int\limits_{0}^{2} r^3 dr \]

Answer 12

\[2 \pi \times \frac{16}{4} = 8\pi \]the second integral is alot easier to compute

you get