Hints only please !! I gave a test today, and it was asked to find area enclosed by the curve $x^4 + y^4 = 2*x*y$. This is an implicit function. A quick obs. shows me that, the curve is entirely bound and (1,1)&(-1,-1) are the interesting points beside origin. To find the area, I attempted $ I = \int y dx $ taking by parts but to no avail as I am neither getting the required limits, nor a well-known function by manipulation. I searched for similar problems; they used polar substitution. I am finding it to be clumsy and maybe not appropriate for this integral. Please give a hint as to how proceed.

  • $\begingroup$ Learn to like the polar coordinate way of doing this. With practice you will become less clumsy. $\endgroup$ – kimchi lover Jan 27 at 16:33
  • $\begingroup$ Can you guide me on finding proper limits in this case? $\endgroup$ – Sarthak Rout Jan 27 at 16:41
  • $\begingroup$ Change variables to $r$ and $\theta$, and find an equation relating them. When I do this I get something of form $r^2=f(\theta)$. Certain values of $\theta$ make $f(\theta)<0$; your range of integration is over those $\theta$ such that $f(\theta)\ge0$. If you believe your sketch of the curve you should be able to read the $\theta$ range off of it, using the relation that $\tan\theta=y/x$. $\endgroup$ – kimchi lover Jan 27 at 16:57

Setting $x=r \cos(\theta), y = r \sin(\theta)$ (the classical change of variables : polar $\to$ cartesian) in your equation gives

$$r^4(\underbrace{(\cos(\theta)^2+\sin(\theta)^2)^2}_{=1}-\underbrace{2\cos(\theta)^2.\sin(\theta)^2}_{\tfrac12 \sin(2 \theta)})=r^2 \underbrace{2 \cos(\theta).\sin(\theta)}_{\tfrac12 \sin(2 \theta)}.$$

Simplifying by $r^2$ and taking the square root, one gets :

$$r(\theta)=\sqrt{\dfrac{\sin(2\theta)}{1-\tfrac12\sin(2\theta)^2}} \ \ \ \text{if} \ \ \ \sin{2 \theta} \geq 0$$

The last condition means that the curve will exist if $0 \leq \theta \leq \pi/2$ (first quadrant) and/or $\pi \leq \theta \leq 3\pi/2$ (third quadrant). No part of the curve in the second and fourth quadrants.

It remains to integrate to obtain the area enclosed by the curve:

$$A=\frac12\int_0^{2 \pi}r(\theta)^2 d \theta$$

BUT, this integral is equal to $0$ because we turn once in the positive sense, once in the negative one. We have to take a serious look at the curve :

enter image description here

We are obliged, if we want to have the unsigned area of a "petal" to integrate from $0$ to $\pi/2$ (and afterwards double the result as a final answer). Up to you for the calculations (as you desire mainly hints).

  • $\begingroup$ This integral is the classical formula for area enclosed by polar curves not passing through the origin. I just modified my text, showing that one has to consider in this case the different quadrants (the first quadrant corresponding in polar coordinates to $0 \leq \theta \leq \pi/2$). The first petal is traversed in the anticlockwise sense whereas the second petal is traversed in the clockwise sense, thus annihilating the first result. $\endgroup$ – Jean Marie Jan 27 at 18:00
  • $\begingroup$ I had reached here, than I substituted $ y = rsin(\theta), x = rcos(\theta)$ in integral of ydx to get a clumsy integral...#Can you please explain why do I need to integrate$ r^2d(\theta) $from 0 to 2pi? I can proceed from here.. The integral is easy.. Kimchi lover says $\theta$ must be such that $f(\theta) $is positive. So$ \theta$ varies from (0 to pi/2 and pi to 3pi/2) ... Is the ans $\pi/2$ ? Please clarify this part only : # $\endgroup$ – Sarthak Rout Jan 27 at 18:01
  • $\begingroup$ Let me search a reference to formula you mentioned.. $\endgroup$ – Sarthak Rout Jan 27 at 18:04
  • $\begingroup$ I computed the area of the first petal : $\pi/4$ (very likely when you see the figure) $\endgroup$ – Jean Marie Jan 27 at 18:04
  • $\begingroup$ Ok, I got my doubt : It is the area of the differential triangle $\endgroup$ – Sarthak Rout Jan 27 at 18:06

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