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Let $\mathcal{C}$ be the curve defined by the vector function $\vec r(t)=(1-t^2)\vec i+(t-t^3)\vec j$ with $t\in \Bbb R$. I need to find the area confined in the closed loop $\gamma$ formed by $\mathcal{C}$, using Green's theorem.

$\gamma$ is smooth, closed, simple and arbitrarily oriented counter clockwise, so :

$$\int\int_DdA=\frac{1}{2}\oint_\gamma xdy-ydx$$

We have : $\begin{cases} x=1-t^2\\ y=t-t^3 \end{cases}$

And by the figure I have of the curve (I don't know how we could plot it otherwise), I can tell that the loop starts at $(0,0)$ and ends at $(1,0)$ so that $t\in[0,1]$.

Which leads to :

$$\int\int_DdA=\frac{1}{2}\oint_{0}^{1} ((1-t^2)(1-3t^2)-(t-t^3)(-2t))dt$$ $$=\frac{1}{2}\oint_{0}^{1} (t^4-2t^2+1)dt=4/15$$

I think this was simple enough and that I got it right. However, now, knowing this result, I must calculate the following integral :

$$\oint_{\gamma}\sqrt{1+x^3}dx+2xdy$$

We know that, as per Green's theorem, it can be rewritten as :

$$\int\int_D\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)dA=\int\int_D(2-0)dA$$

Sweet, the root vanishes ! But I don't know how to define domain $D$.

Here is a figure of domain $D$ :

Domain to integrate over.

Any help would be much appreciated. Thanks !

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There are two values of t that give the origin for that parameterization: t =-1 and t =1. I believe if you just evaluate the line integral with those limits you'll get the area inside the closed part of the curve.

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  • $\begingroup$ Thanks, indeed. But my question really is how to define the domain inside the loop in cartesian form to be able to solve the last integral. $\endgroup$ – Bob Leponge Jul 20 '15 at 0:59
  • $\begingroup$ I see. Eliminating t from the equation of the curve should give $y=x\sqrt{1-x}$ and $y=-x\sqrt{1-x}$. It's easy to do this if you notice that $\frac{y}{x}=t$ $\endgroup$ – Alex Pavellas Jul 20 '15 at 1:13
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We have the curve $C$ defined parametrically as

$$\begin{align} \vec r(t)&=\hat xx(t)+\hat yy(t)\\\\ &=\hat x(1-t^2)+\hat yt(1-t^2)\\\\ &\implies x(t)=1-t^2\,\,\text{and}\,\,\,y(t)=t(1-t^2)\tag 1 \end{align}$$

for $t\in [0,1]$.


NOTE:

For $t\in[0,1]$, we see that $y$ is greater than $0$. The closed curve $\gamma$ is formed by two parts; (i) the curve $C$ defined by the aforementioned parameterization and (ii) the line segment $C_x$ along the x-axis from $(0,0)$ to $(1,0)$. However, we note that on the line segment, $y=0$ and $dy=0$ and the contribution from the integration over $C_x$ is zero. We therefore have$$\oint_{\gamma}(xdy-ydx)=\int_{C}(xdy-ydx)+\int_{C_x}(xdy-ydx)=\int_{C}(xdy-ydx)$$


Now, using the first relationship in $(1)$ we find that

$$t^2=1-x \tag 2$$

while using both relationships in $(1)$ simultaneously we find that

$$t^2=(y/x)^2 \tag 3$$

Noting that the right-hand sides of $(2)$ and $(3)$ are equal reveals

$$y^2=x^2(1-x)$$

We recall that $t\in[0,1] \implies x\in[0,1]\,\,\text{and}\,\,y\ge0$.

Thus, the area $A$ inside the region bounded by $y=0$ and $y= x\sqrt{1-x}$ is given by

$$\begin{align} A&=\int_0^1\int_{0}^{x\sqrt{1-x}}dydx\\\\ &=\int_0^1\,x\sqrt{1-x}\,dx\\\\ &=-\frac{2}{3}\left.x(1-x)\right|_0^1+\frac23\int_0^1\,(1-x)^{3/2}\,dx\\\\ &=-\frac23\,\frac{2}{5}\,\left.(1-x)^{5/2}\right|_0^1\\\\ &=\frac{4}{15} \end{align}$$

as expected!!

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  • $\begingroup$ @bobleponge I included explanation of a subtlety in the problem along with a full development of the evaluation of the integral for the area. Please let me know how I can improve my answer. I really just want to give you the very best answer I can. $\endgroup$ – Mark Viola Jul 22 '15 at 15:41

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