Area of loop in x-y plane

[littlemathquark]

Homework Statement

Find the area of loop of $x^5+y^5=3x^2y^2$

Relevant Equations

Find the area of loop of $x^5+y^5=3x^2y^2$

Let $x=rcos\theta$, $y=rsin\theta$ and area=$(1/2)\int_{\theta_1}^{\theta_2}r^2.d\theta=$ $(3/8)\int_0^{\pi/2}{(((sin 2x)^2)^2 /((cos x)^5+(sin x)^5)^2}) d\theta$
İf it is ok I go on.

 

[Mark44]

Homework Statement: Find the area of loop of $x^5+y^5=3x^2y^2$
Relevant Equations: Find the area of loop of $x^5+y^5=3x^2y^2$

Let $x=rcos\theta$, $y=rsin\theta$ and area=$(1/2)\int_{\theta_1}^{\theta_2}r^2.d\theta$= $(3/4)\int_0^{\pi/2}{((sin 2x)^2)^2 /((cos x)^5+(sin x)^5)^2}$
İf it is ok I go on.

Hard to say whether you're on the right track without knowing what the curve looks like, whether the limits of integration you show, or whether your integrand is reasonable for this problem. There's a lot you didn't show here.

BTW, I fixed your broken LaTeX. You might want to check it to be sure it's what you intended.

 

[littlemathquark]

İt looks folium of Descartes I mean $x^3+y^3=3axy$

 

[littlemathquark]

This is the curve $x^5+y^5=3x^2.y^2$

 

[littlemathquark]

From given equation if $x=rcos\theta$ and $y=rsin\theta$ then $r=\dfrac{3cos^2\theta.sin^2\theta} {cos^5\theta +sin^5\theta }$

 

[Mark44]

From given equation if $x=rcos\theta$ and $y=rsin\theta$ then $r=\dfrac{3cos^2\theta.sin^2\theta} {cos^5\theta +sin^5\theta }$

Yes, this looks OK for r, but the integral looks quite complicated.

 

[pasmith]

From given equation if $x=rcos\theta$ and $y=rsin\theta$ then $r=\dfrac{3cos^2\theta.sin^2\theta} {cos^5\theta +sin^5\theta }$

I believe this simplifies to <br /> r = \frac{3 \sec \theta \tan^2 \theta}{1 + \tan^5 \theta} which suggests the substitution u = \tan \theta.

 

[littlemathquark]

I believe this simplifies to <br /> r = \frac{3 \sec^3 \theta \tan \theta}{1 + \tan^5 \theta} which suggests the substitution u = \tan \theta.

İt must be a mistake.If numerator and denumarator product by $1/cos^5\theta$
$r=\dfrac{3sec^3\theta.sin^2\theta } {1+tan^5\theta}$

 

[PeroK]

İt must be a mistake.If numerator and denumarator product by $1/cos^5\theta$
$r=\dfrac{3sec^3\theta.sin^2\theta } {1+tan^5\theta}$

He was testing you!

 

[littlemathquark]

I believe this simplifies to <br /> r = \frac{3 \sec \theta \tan^2 \theta}{1 + \tan^5 \theta} which suggests the substitution u = \tan \theta.

Thank you but I think it's easier product numerator and denominator of $r^2$ by $1/cos^{10} \theta$ and substitution $u=1+tan^5\theta$

 

[Mark44]

İt must be a mistake.If numerator and denumarator product by $1/cos^5\theta$
$r=\dfrac{3sec^3\theta.sin^2\theta } {1+tan^5\theta}$

Here's a tip for you: Instead of using a period to indicate multiplication, use \cdot.
Like this: $r=\dfrac{3sec^3\theta \cdot \sin^2\theta } {1+tan^5\theta}$

Also, instead of "if numerator and denumerator product by ..." write "if numerator and denominator are multiplied by ..."

In English, the expression at the bottom of a fraction is called the denominator. As far as I know, denumerator isn't a word in English.

 

[littlemathquark]

Here's a tip for you: Instead of using a period to indicate multiplication, use \cdot.
Like this: $r=\dfrac{3sec^3\theta \cdot \sin^2\theta } {1+tan^5\theta}$

Also, instead of "if numerator and denumerator product by ..." write "if numerator and denominator are multiplied by ..."

In English, the expression at the bottom of a fraction is called the denominator. As far as I know, denumerator isn't a word in English.

Thank you, it"s typo.

 

[littlemathquark]

$f(x, y) =x^5+y^5-3x^2y^2=f(y,x)=0$

$x=rcos\theta$ and $y=rsin\theta$ $r=\dfrac {3cos^2\theta. sin^2\theta } {cos^5\theta +sin^5\theta }$

$E^*=\{(r, \theta) :0\le r\le \dfrac {3cos^2\theta sin^2\theta } {cos^5\theta +sin^5\theta }, 0\le\theta \le\pi/4\}$

$S(E) =2\iint\limits_{E^*} rdrd\theta=\int_0^{\pi/4}(\dfrac{3cos^2\theta. sin^2\theta }{cos^5\theta +sin^5\theta }) ^2d\theta =\int_0^{\pi/4}\dfrac{9cos^4\theta. sin^4\theta}{(cos^5\theta +sin^5\theta) ^2}d\theta$
Dividing numerator and denominator by $cos^{10}\theta$
$=9.\int_0^{\pi/4}\dfrac{tan^4\theta. sec^2\theta }{(1+tan^5\theta) ^2}d\theta$
$u=1+tan^5\theta$
$=\dfrac 95\int_1^2 \dfrac{du} {u^2}=0,9$

 

[SammyS]

İt must be a mistake.
If numerator and denominator product are multiplied by $1/cos^5\theta$
$r=\dfrac{3\sec^3\theta\cdot\sin^2\theta } {1+\tan^5\theta}$

(I edited the above for readability.)

Notice that @pasmith made a correction to his post about an hour after you posted the above.

His corrected result: $\displaystyle \quad r = \frac{3 \sec \theta \tan^2 \theta}{1 + \tan^5 \theta}$

This is equivalent to your expression.

Were you able to find the area of the loop?

What did you get for the area?

 

[littlemathquark]

Yes, I found 0,9 for the area of loop and I wrote my solution at my
last(13) massage.

 

[SammyS]

Yes, I found 0,9 for the area of loop and I wrote my solution at my
last(13) massage.

It looks good to me.

(I was taking my time with writing up my previous post. At the time I started writing my post, your solution wasn't yet posted.)

 

[littlemathquark]

Are there a different solution?

 

[fresh_42]

Are there a different solution?

We could rotate the curve by $-45°$ so that it becomes a function, say $v=v(u).$ The result would then be
$$
2\int_0^3 v\,du=2\int_0^{2,12132} v(u)\,du
$$
but $v(u)$ is a nightmare:
$$
v(u)=\sqrt{
-\dfrac{5\sqrt{2}u+3}{5\sqrt{2}u-3}+\dfrac{2}{5\sqrt{2}u-3}\sqrt{10u^2+12\sqrt{2}u}}
$$

 

[renormalize]

Are there a different solution?

Not really a "different" solution, but here's a useful equivalent way to set up the general problem of finding the area $A$ of any 2D region $R$ bounded by the closed-contour $\partial R$ by using Green's theorem:$$\oint_{\partial R}\left(Pdx+Qdy\right)=\iint_{R}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dxdy\tag{1}$$Set $Q=x,P=0$ or $Q=0,P=y\tag{1}$ to find:$$A\equiv\iint_{R}dx\,dy=\oint_{\partial R}x\,dy=-\oint_{\partial R}y\,dx=\frac{1}{2}\oint_{\partial R}\left(x\,dy-y\,dx\right)=\frac{1}{2}\intop_{0}^{\theta_{\text{max}}}\left[r\left(\theta\right)\right]^{2}d\theta\tag{2}$$where in the final form we've switched to polar coordinates using $x=r\left(\theta\right)\cos\theta,y=r\left(\theta\right)\sin\theta$, and $\theta_{\text{max}}$ is chosen so that the integration range sweeps exactly once around the contour. For your specific problem eq.(2) becomes:$$A=\frac{1}{2}\intop_{0}^{\pi/2}\left[\frac{3\cos^{2}\theta\sin^{2}\theta}{\cos^{5}\theta+\sin^{5}\theta}\right]^{2}d\theta=\frac{9}{10}\tag{3}$$which is exactly your result.

 

[littlemathquark]

We could rotate the curve by $-45°$ so that it becomes a function, say $v=v(u).$ The result would then be
$$
2\int_0^3 v\,du=2\int_0^{2,12132} v(u)\,du
$$
but $v(u)$ is a nightmare:
$$
v(u)=\sqrt{
-\dfrac{5\sqrt{2}u+3}{5\sqrt{2}u-3}+\dfrac{2}{5\sqrt{2}u-3}\sqrt{10u^2+12\sqrt{2}u}}
$$

We could rotate the curve by $-45°$ so that it becomes a function, say $v=v(u).$ The result would then be
$$
2\int_0^3 v\,du=2\int_0^{2,12132} v(u)\,du
$$
but $v(u)$ is a nightmare:
$$
v(u)=\sqrt{
-\dfrac{5\sqrt{2}u+3}{5\sqrt{2}u-3}+\dfrac{2}{5\sqrt{2}u-3}\sqrt{10u^2+12\sqrt{2}u}}
$$

How can you find this function? Can you give some detail?

 

[fresh_42]

How can you find this function? Can you give some detail?

I have deleted my scribblings since then. I used WA to picture the graph (as in post #4) and saw the symmetry at 45°. Then apply the rotation matrix to (x,y) and get (u,v). This gives a polynomial in (u,v). I guess I used WA again to resolve it for v. If we demand v>0 then we get half the area, but we gain a function v(u) instead of only a relation.