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**Dr Zoidburg** Question:

Use Green's theorem to evaluate the following line integrals. Assume that the curve is transversed counterclockwise.

(a) $\displaystyle \oint_{\gamma}3xy.\partial x + 2xy.\partial y, $ where $\displaystyle \gamma $ is the rectangle with corners at (-2,1),(-2,2),(4,1) & (4,2).

(b)$\displaystyle \oint_{\gamma}(x^2 - y).\partial x + x.\partial y, $ where $\displaystyle \gamma $ is the circle $\displaystyle x^2+y^2=4$

My answers:

(a) P=3xy & Q=2xy

Green's Theorem gives:

$\displaystyle \int \int_{D}(2y-3x).\partial A$

where D is the rectangle enclosed by $\displaystyle -2 \leq x \leq 4 \; and \; 1 \leq y \leq 2 $

I found if I just go ahead and integrate I end up with 0 as the answer.

However, since the path is not smooth, should I take the integral along each side separately and then add them?

Doing this I end up with:

first side (-2,1) to (-2,2). Make x=-2 -> dx = 0. Make y=t -> dy=dt.

$\displaystyle \int_{(-2,1)}^{(-2,2)}3xy.\partial x + 2xy.\partial y = \int_{t=1}^{2}3(-2)t.0 + 2(-2)t.\partial t = \int_{1}^{2}-4t.\partial t = \bigg[ -2t^2 \bigg]_{1}^{2} = -6$

I did the same for the other 3 sides and got 12, 18 and 36. This makes the entire integral 60.

Does this look ok? Or do I need to take the absolute values of each, making the final answer 72.

(b) $\displaystyle P=x^2-y \; and \; Q=x$

Green's theorem gives:

$\displaystyle \oint_{\gamma } (x^2-y).\partial x + x.\partial y = \int \int_{D}2.\partial A$

where D is a disk of radius 4 centred at the origin.

Since this is a disk, is the best way to do this integral is using polar co-ordinates?

This is what I got:

$\displaystyle \int_{0}^{2\pi} \int_{0}^{4}2r.\partial r \partial \theta = \int_{0}^{2 \pi}\bigg[r^2\bigg]_{0}^{4}=\int_{0}^{2\pi}16.\partial\theta =\bigg[16\theta \bigg]_{0}^{2\pi}=32\pi $