I have the equation of a curve

$x^{2/3} + y^{2/3} = a^{2/3}$

and I'm using the change of variables

$x = u\cos^3v $, $y = u\sin^3v$ .

I have calculated the Jacobian $\frac{\partial(x,y)}{\partial(u,v)}$, but how do I use all this stuff to find the area of the region bounded by the curve and the positive x- and y-axes? I'm having particular difficulty figuring out the new limits for the double integral wrt u and v.


  • $\begingroup$ Can you use Green's Theorem? $\endgroup$ Feb 3 '14 at 21:34
  • $\begingroup$ @Fantini I don't think so :( $\endgroup$
    – Taimur
    Feb 3 '14 at 21:34
  • $\begingroup$ Instead of $u$ you should have $a$ and then things are straightforward. $\endgroup$ Feb 3 '14 at 21:55
  • $\begingroup$ @MhenniBenghorbal I was given the change of variables in the question :( $\endgroup$
    – Taimur
    Feb 3 '14 at 21:57
  • $\begingroup$ It sounds like a mistake on the teachers part. $\endgroup$ Feb 3 '14 at 21:57

Why not just solve for $y$ in terms of $x$ and then integrate over $x \in [0,a]$?

$$y = \left ( a^{2/3}-x^{2/3} \right )^{3/2} = a \left [ 1-\left ( \frac{x}{a} \right )^{2/3} \right ]^{3/2}$$

Let $x=a \sin^3{t}$; then the integral is

$$3 a^2 \int_0^{\pi/2} dt \, \cos^4{t} \, \sin^2{t} = \frac{3}{8} a^2 \int_0^{\pi/2} dt \, (\sin^2{2 t} + \cos{2 t} \sin^2{2 t}) = \frac{3 \pi}{32} a^2$$


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