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$$ \gamma (t) = \cos (t) \; \vec{i} + sin (t) \; \vec{j} $$ with $ \;\; t=0 \ldots\pi $

first using $x$ as a parameter and then $y$ as a parameter.

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  • $\begingroup$ By $t=0\ldots\Pi$ I assume you mean $t\in[0,\pi]$? $\endgroup$
    – Will R
    Dec 19, 2016 at 11:10

1 Answer 1

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I don't understand what does "using x and then y as a parameter" mean in this case. Here, you're going to get minus twice the area of the domain determined by $\;\gamma\;$ together with the $\;x\,-$ axis, i.e.: the area of a unit circle (this follows from Green's theorem "closing" the curve):

$$\int_Cydx-xdy=\int_0^\pi\left[\sin t(-\sin t)-\cos t(\cos t)\right]dt=-\int_0^\pi1\,dt=-\pi$$

Using Green's Theorem with the vector field

$$F(x,y)=\left(xy,\,xy\right)\implies\iint_B\left(y-x\right)dA=\int_Cydx-xdy+\int_{C_1}ydx-xdy$$

where $\;C_1: (t,0)\;,\;\;-1\le t\le1\;$ . But

$$\int_{C_1}ydx-xdy=\int_{-1}^1(0\cdot1-t\cdot 0)dt=0$$

and thus the result

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  • $\begingroup$ I'm sorry.I mean first make an integral considering x as a parameter. Do the same recital and as parameter. But what you did confers on the response my teacher gave, I just did not understand how to build that result. $\endgroup$ Dec 19, 2016 at 11:29

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