0
$\begingroup$

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 $dxdy$.

$\endgroup$
  • 1
    $\begingroup$ It means $(x - y)(dx + dy) = (x-y)dx + (x-y)dy$. $\endgroup$ – IAmNoOne May 7 at 3:00
0
$\begingroup$

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)

enter image description here 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)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – nestealova May 7 at 3:22
  • $\begingroup$ Yes I've edited it $\endgroup$ – Ak19 May 7 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.