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Use greens theorem to calculate $\int_{\gamma}y\,dx+x^2dy$ where $\gamma$ is the following closed path: (a) The circle given by $g(t) = (\cos(t), \sin(t)), 0 \le t \le 2\pi$.

What I have tried:

Using the following $$\int_\gamma P\,dx+Q\,dy = \int_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)$$

I have that $$\frac{\partial (x^2)}{\partial x}-\frac{\partial (y)}{\partial y} = 2x-1$$

Replacing $x = \cos(t)$ I then have:

$$\int_0^{2\pi}(2\cos(t)-1)\,dt=-2\pi$$

But the answer in my book show $-\pi$, so I have tried replacing $x = \cos(t), y = \sin(t), dx = -\sin(t), dy = \cos(t)$ in the integral but that gives me $\pi$. Have I made a mistake?

Bernard
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me.limes
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2 Answers2

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As you are applying Green's theorem instead of direct line integral, you are supposed to integrate over the area of the closed region.

$ \displaystyle \int_\gamma Pdx+Qdy = \iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) ~ dA$

As the region is unit circle, you should use $x = r \cos t, y = r \sin t$, $0 \leq r \leq 1, 0 \leq t \leq 2\pi$

So the integral is,

$ \displaystyle \int_0^{2\pi} \int_0^1r (2r\cos t - 1) ~ dr ~ dt$

Math Lover
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    Also note that $x$ is an odd function so its integral over the unit circle will be zero due to symmetry of the region about y-axis. So all you are left with is to integrate $ - \int_0^{2\pi} \int_0^1 r~ dr ~ dt = - \pi$c, which is area of the unit circle with negative sign – Math Lover Nov 06 '21 at 17:01
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A mistake is that you confused the calculations of double integrals and line integrals.

You are supposed to calculate the double integral $\iint_D (2x-1)\;dxdy$. The "replacing $x=\cos(t)$ step does not make sense.

Here $D$ is the set $\{(x,y):x^2+y^2\le 1\}$. One should use polar coordinate to get $$ \iint_D (2x-1)\;dxdy=2\iint_D x\;dxdy-\iint_D 1\;dxdy =2\int_0^{2\pi}\int_0^1(r\cos\theta) r\;drd\theta-\pi=0-\pi $$