# Area of region

1. Jan 20, 2014

### etf

Here is my task and attempt for solution:

As you can see, I tried with substitutions but I have no idea what to do next.
Any suggestion?

Last edited: Jan 20, 2014
2. Jan 21, 2014

### haruspex

The equation you start with looks nasty. Are you sure it's not supposed to be x3+8y3=x2+4y2, or somesuch?
Either way, I can't see the merit of that substitution. Why not just go to the usual polar coordinates? At least you can get as far as writing the area as a single integral. Still tough though.

3. Jan 21, 2014

### etf

I'm sure it's how I wrote it... I tried usual polar coordinates too and I got that r=(4*cos(phi)^2+sin(phi)^2)/(cos(phi)^3+8*sin(phi)^3). I don't think that this substitution is more useful than previous. I tried with few other substitutions in order to make region easier but without success. I really have no idea what to do here :(

Last edited: Jan 21, 2014
4. Jan 21, 2014

### haruspex

I agree it's still nasty, but at least you can write out the integral for the area now . With your other substitution you'd have to calculate quite a messy Jacobian to get that far. (Did you compute the Jacobian? Maybe it simplifies magically, but I doubt it.)

5. Jan 22, 2014

### haruspex

I've had maybe a better idea. Try u = x, v = y/x.

6. Jan 23, 2014

### etf

Wow! This one is key to my problem :)
If I didn't make mistake somewhere, area should be:

7. Jan 23, 2014

### haruspex

Looks good until you get to the last line. A = ∫∫dxdy = ∫∫Jdudv etc. I'm afraid it's going to get quite complicated, but it's all doable.

8. Jan 23, 2014

### etf

I computed it using Matlab, I didn't try to do it by hand...
Thanks a lot!

9. Jan 23, 2014

### haruspex

OK, I see - I misread your double integral as two separate integrals. (That's not the way I write them.)
The simplicity of the answer does suggest we've missed a trick somewhere.
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