# How to evaluate integral of (x - y)(dx + dy) with Green's Theorem?

I want to evaluate the integral $$\int(x - y)(dx + dy)$$ along curve C where C is the semicircular part of $$x^2 + y^2 = 4$$ above $$y = x$$ from $$(-\sqrt2, -\sqrt2)$$ to $$(\sqrt2, \sqrt2)$$ using Green's Theorem. What is meant by $$(x - y)(dx + dy)$$? Usually it is in the format $$dxdy$$.

• It means $(x - y)(dx + dy) = (x-y)dx + (x-y)dy$. – IAmNoOne May 7 at 3:00

By Green's theorem,

$$I = \int_c (Pdx+Qdy) = \int\int{\bigg(\frac{\partial Q}{\partial x}} - \frac{\partial P}{\partial y}\bigg)dx\ dy$$

Here $$P = Q = x-y$$

$$Q_x = 1 \ , P_y = -1$$, $$Q_x - P_y = 2$$

$$I = \int\int_{s_-}2dxdy$$ ( - for clockwise direction)

Now from the graph,

Area of the semicircle (anticlockwise), $$\int\int_{s_+}dxdy = \frac{\pi}{2} r^2 = \frac{\pi}{2}(4) = 2\pi$$

Thus $$I = 2\int\int_{s_-}dxdy = -2.2\pi = -4\pi$$ (clockwise)

• I think since it goes from $(-\sqrt(2), -\sqrt(2)) to (\sqrt(2), \sqrt(2))$ it is in clockwise direction and so result would be negative. – nestealova May 7 at 3:22
• Yes I've edited it – Ak19 May 7 at 4:09