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Let $R$ be the region enclosed by the curve parameterized by $g(t) = (t^4 - t^2, t^6 - t^2)$, where $0 \leq t \leq 1.$ Find the area of $R$ using Green's Theorem. In order to use the green's theorem, my text book says we need a vector field but this question does not provide the vector field. How can I approach this??

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  • $\begingroup$ Pick your field $\langle P,Q \rangle$, so that $\partial Q/\partial x - \partial P/\partial y = 1$. $\endgroup$ – Mark McClure Dec 22 '14 at 4:57
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Since $R$ is a simply connected region bounded by the curve $g$, Green's Theorem tells you that $$\displaystyle\iint\limits_{R}\left[\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right]\,dx\,dy = \oint\limits_{g}P\,dx+Q\,dy$$

for functions $P(x,y)$ and $Q(x,y)$.

You want to compute the area of $R$, which is given by $\displaystyle\iint\limits_{R}1\,dx\,dy$.

To make the formula for Green's Theorem useful for calculating the area of $R$, you should pick functions $P(x,y)$ and $Q(x,y)$ such that $\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y} = 1$. One such choice is $P(x,y) = 0$ and $Q(x,y) = x$. Another choice is $P(x,y) = -y$ and $Q(x,y) = 0$. Yet another choice is $P(x,y) = -\frac{1}{2}y$ and $Q(x,y) = \frac{1}{2}x$.

This gives you the following formulas for the area of $R$: $$\displaystyle\iint\limits_{R}1\,dx\,dy = \oint\limits_{g}x\,dy = -\oint\limits_{g}y\,dx = \dfrac{1}{2}\oint\limits_{g}x\,dy-\dfrac{1}{2}\oint\limits_{g}y\,dx$$

Any one of those last three line integrals can be used to compute the area of $R$.

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I found $\int\limits^1_0 t^5(t^2-1)^2 dt = \frac{1}{60}$ with the same use of the theorem than JimmyK4542

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