Can we find the interior of the given curve using Green's theorem?

[evinda]

Hello! (Wave)

Using Green's theorem, I want to compute the integral

$$\oint_C ydx+xdy$$

where $C$ has the parametric representation $r(t)=2 \cos^3 t i+ 2 \sin^3 t j, (0 \leq t \leq 2 \pi)$.

Using Green's theorem, we get that $\oint_C ydx+xdy=\iint_U (1-1)dxdy=0$.

I am wondering if we could find the interior $U$ of the curve $C$. I.e. if the result wouldn't be $0$, could we compute the double integral that we get? (Thinking)

 

[I like Serena]

Hey evinda! (Smile)

To calculate the integral we can either try to set it up in polar or in cartesian coordinates.

To get cartesian coordinates we have in the first quadrant:
$$x=2\cos^3t \quad\Rightarrow\quad
t=\arccos \left((x/2)^{1/3}\right) \quad\Rightarrow\quad \\
y=2\sin^3 t = 2\sin^3\left(\arccos \left((x/2)^{1/3}\right)\right) = 2\left(\sqrt{1-(x/2)^{2/3}}\right)^3
= 2\left(1-(x/2)^{2/3}\right)^{3/2}
$$
Note that $U$ has horizontal and vertical reflection symmetry.
So:
$$
\iint_U f(x,y) dxdy = \int_0^2 \int_{-2\left(1-(x/2)^{2/3}\right)^{3/2}}^{2\left(1-(x/2)^{2/3}\right)^{3/2}} f(x,y)dydx + \text{almost same integral for negative x}
$$
If $f(x,y)=1$, Wolfram can come up with a closed form.
See here.
It's not a particularly easy integral to find by hand, but it seems doable.

And of course it changes if $f(x,y)$ is not a constant, in which case it can become easier or more complicated. (Thinking)

 

[evinda]

Using the fact that

$$\sin^2\left(\arccos \left(\left(\frac{x}{2}\right)^{\frac{1}{3}}\right)\right)+\cos^2\left(\arccos\left(\left(\frac{x}{2}\right)^{\frac{1}{3}}\right)\right)=1$$

we get that

$$\sin\left(\arccos\left(\frac{x}{2}\right)^{\frac{1}{3}}\right)=\pm \sqrt{1-\left(\frac{x}{2}\right)^{\frac{2}{3}}}$$

So it follows that $y=\pm 2 \left( 1-\left( \frac{x}{2}\right)^{\frac{2}{3}}\right)^{\frac{3}{2}}$.

Since $1-\left( \frac{x}{2}\right)^{\frac{2}{3}}$ is an expression under the square root, it must be positive, and so it follows that $x \leq 2$. How do we get that $x \geq 0$ ? (Thinking)

 

[I like Serena]

Since $1-\left( \frac{x}{2}\right)^{\frac{2}{3}}$ is an expression under the square root, it must be positive, and so it follows that $x \leq 2$. How do we get that $x \geq 0$ ? (Thinking)

It's just that $x^{2/3}$ is not defined for negative $x$. (Worried)

Then again, if we write it as $(x^2)^{1/3}$ it would be fine. (Thinking)

 

[evinda]

It's just that $x^{2/3}$ is not defined for negative $x$. (Worried)

Why isn't $x^{2/3}$ defined for negative $x$? (Thinking)

 

[I like Serena]

Why isn't $x^{2/3}$ defined for negative $x$? (Thinking)

Behold:
$$-1=(-1)^{\frac 23 \cdot \frac 32} = \left((-1)^{\frac 23}\right)^{\frac 32} = 1^{\frac 32} = 1$$
yes?
... but isn't $-1\ne 1$? (Sweating)

Can you spot the problem? (Wondering)

 

[evinda]

Behold:
$$-1=(-1)^{\frac 23 \cdot \frac 32} = \left((-1)^{\frac 23}\right)^{\frac 32} = 1^{\frac 32} = 1$$
yes?
... but isn't $-1\ne 1$? (Sweating)

Can you spot the problem? (Wondering)

We have that $x=x^{\frac{2}{3} \cdot \frac{3}{2}}=(x^{\frac{3}{2}})^{\frac{2}{3}}$.

Since $x^{\frac{3}{2}}=x \sqrt{x}$, $x$ has to be $\geq 0$, right?

 

[I like Serena]

We have that $x=x^{\frac{2}{3} \cdot \frac{3}{2}}=(x^{\frac{3}{2}})^{\frac{2}{3}}$.

Since $x^{\frac{3}{2}}=x \sqrt{x}$, $x$ has to be $\geq 0$, right?

If we would evaluate the expression in that order, then yes, $x$ has to be $\geq 0$.
But in the order I wrote it, that is not required, is it? (Wondering)
Either way, yes, we can see that the order in which we evaluate the powers makes a difference.

To be fair, we can define $x^{2/3}$ for negative $x$, so that $(-x)^{2/3}=x^{2/3}$.
It's just that if do, we can no longer use the power identity $a^{bc}=(a^b)^c$ as it breaks down.
It's one or the other.
So generally $x^\alpha$ is not defined for negative $x$ if $\alpha$ is a rational or real number that is not an integer. (Nerd)

Just for fun, let's see what W|A makes of it:
(-1)^(2/3) - Wolfram|Alpha Results
Hey, where did that come from? (Wondering)

 

[evinda]

To be fair, we can define $x^{2/3}$ for negative $x$, so that $(-x)^{2/3}=x^{2/3}$.
It's just that if do, we can no longer use the power identity $a^{bc}=(a^b)^c$ as it breaks down.
It's one or the other.

I am confused now. Doesn't this property always hold? (Thinking)

 

[I like Serena]

I am confused now. Doesn't this property always hold? (Thinking)

Indeed, that power identity does not always hold.
See the section Failure of power and logarithm identities in the Exponentation wiki article to see how it breaks down for complex numbers. (Worried)

 

[evinda]

$$
\iint_U f(x,y) dxdy = \int_0^2 \int_{-2\left(1-(x/2)^{2/3}\right)^{3/2}}^{2\left(1-(x/2)^{2/3}\right)^{3/2}} f(x,y)dydx + \text{almost same integral for negative x}
$$

Here, how did you find this one: $\int_0^2 \int_{-2\left(1-(x/2)^{2/3}\right)^{3/2}}^{2\left(1-(x/2)^{2/3}\right)^{3/2}} f(x,y)dydx + \text{almost same integral for negative x}$ ? (Worried)

I mean, isn't there a formula that holds for all $x$ ? Instead of considering some integral for negative $x$ ?

 

[I like Serena]

Here, how did you find this one: $\int_0^2 \int_{-2\left(1-(x/2)^{2/3}\right)^{3/2}}^{2\left(1-(x/2)^{2/3}\right)^{3/2}} f(x,y)dydx + \text{almost same integral for negative x}$ ? (Worried)

I mean, isn't there a formula that holds for all $x$ ? Instead of considering some integral for negative $x$ ?

Yes, I believe we can make it:
$$\int_{-2}^2 \int_{-2\left(1-((x/2)^2)^{1/3}\right)^{3/2}}^{2\left(1-((x/2)^2)^{1/3}\right)^{3/2}} f(x,y)dydx$$
I just tried to avoid that profusion of parentheses. (Wasntme)

 

[evinda]

Yes, I believe we can make it:
$$\int_{-2}^2 \int_{-2\left(1-((x/2)^2)^{1/3}\right)^{3/2}}^{2\left(1-((x/2)^2)^{1/3}\right)^{3/2}} f(x,y)dydx$$
I just tried to avoid that profusion of parentheses. (Wasntme)

From which relation do we get that $x \geq -2$? (Thinking)

 

[I like Serena]

From which relation do we get that $x \geq -2$?

We have $x=2\cos^3 t$, meaning that $-2 \le x \le 2$. (Thinking)

 

[evinda]

We have $x=2\cos^3 t$, meaning that $-2 \le x \le 2$. (Thinking)

I see... Thanks a lot! (Smirk)