Find area of this bounded curve. I don't know how to start? This question was asked in Mathematical belarusian olympiad.

Question

Find the area bounded by the curve: $$\left(\frac x2 + \frac y3\right)^4= 4xy$$

Tried using subtitutions but couldn't figure it out.

Answer 1

With the change of variable:

$$X=x/2,Y=y/3. \tag{1}$$

the initial equation becomes :

$$(X+Y)^4=24XY \tag{2}$$

Let us compute the area $S$ enclosed by curve $(C)$ defined by equation (2), knowing that finally, when returning to the original variables, you will have to multiply $S$ by $2 \times 3 = 6$.

Some remarks about curve $(C)$ : it is in particular symmetrical with respect to the origin because (2) is invariant by transformation $(X,Y) \to (-X,-Y)$ ; furthermore, $(C)$ is not situated in quadrants II and IV (it is impossible for example that $X<0$ and $Y>0$ simultaneously).

Set $$Y=aX \ \ \ \text{with} \ \ a>0\tag{4}$$

in equation (2) (it amounts to look for the intersection of the curve with straight line having equation (4)).

You will obtain

$$X=\frac{\sqrt{24a}}{(1+a)^2}. \tag{5}$$

As a consequence,

$$Y=aX=a\frac{\sqrt{24a}}{(1+a)^2}\tag{6}.$$

With (5) and (6) we have a parametric representation of curve $(C)$.

Then apply formula for the area of a parametric curve:

$$S=\int_{a=0}^{a=\infty}\frac12(XdY-YdX)$$ for the area of one loop.

Multiply this result by $2$ to account for the second loop, in quadrant III. And don't forget to multiply the result by $6$ as said upwards.

Remark: curve $(C)$ is a lemniscate of Gerono, not to be confused with a lemniscate of Bernoulli.

It can be shown by using a $\pi/4$ coordinates rotation

$$\begin{cases}X+Y&=&\sqrt{2}u\\X-Y&=&\sqrt{2}v\end{cases}$$

transform (2) into the following equation in variables $u,v$:

$$u^4=3(u^2-v^2)$$

which is one of the forms of the equation of such a lemniscate (see this site).

Answer 2

Using GeoGebra, the graph happens to have two loops; one in the first quadrant and the other in the third quadrant.

Substitute : $$x=2r\cos^2a , y=3r\sin^2a$$ On simplifying, we get: $$r^2=6\sin^2(2a)$$

Now consider a sector of angle $da$ making an angle $a$ with $X$ axis.

The area of sector is $$\frac 12 r^2da$$ Therefore, the area of the loop in first quadrant is $$\int_0^{\pi/2}\frac12r^2da$$

Now, the area of both the loops will be twice this integral.

Substituting the value of $r^2$ in the given integral, the area can be found.