# Compute the area under the curve

1. Dec 26, 2009

### (@apache@)

1. The problem statement, all variables and given/known data
Hi everyone,
can you give me some hint to solve this? please

Compute the area under the curve:

2. Relevant equations
$$(x^3+y^3)^2=x^2+y^2$$

3. The attempt at a solution
I've been trying to transform it into polar coordinates, but I've finished when I get the equation for the radius:
$$r^2=1/((\cos(\theta))^3+(\sin(\theta))^3)$$,
because then I get the integral which should not be easily integrated (of course I've compute with Jacobian)

thanks for yout attention

2. Dec 27, 2009

### LCKurtz

Re: Area

Have you looked at the graph? It isn't even clear what region the "area under the curve" describes.

3. Dec 28, 2009

### (@apache@)

Re: Area

I am sorry, I forgot add this condition: $$x>0, y>0$$, so then I will get the integral in this case:
$$\int\limits_{0}^{\pi/2}(\int\limits_{0}^{\frac{1}{\sqrt{\cos^3\alpha+\sin^3\alpha}}}r\,dr)\,d\alpha$$,
which results to $$1/3 (\pi+2 \sqrt{2} \tanh^{-1}(1/\sqrt{2}))$$, but it's little bit different form the correct result which is $$\frac{\pi}{6}+\frac{\sqrt{2}}{3}\ln(1+\sqrt{2})$$

4. Dec 28, 2009

### LCKurtz

Re: Area

Your integral looks OK but your answer is off by a factor of 1/2. You probably dropped it somewhere in the calculation.

5. Dec 28, 2009

### (@apache@)

Re: Area

Yes, you're right. l forget 1/2 in front of integral, then the results are same:
1.8781645380502384306795745680939249649333168829204582775232678988974988081580885734995703727615411934.
Thanks for your time.