# Math Help - difficult integral

1. ## difficult integral

Find the area bounded by the curve given by the equation:
$(x^2 + y^2)^2 = 2a^2(x^2 - y^2)$
How do you transform the equation to the integral?

2. Originally Posted by kezman
Find the area bounded by the curve given by the equation:
$(x^2 + y^2)^2 = 2a^2(x^2 - y^2)$
How do you transform the equation to the integral?
Solve for y:
$(x^2 + y^2)^2 = 2a^2(x^2 - y^2)$

$x^4 + y^4 + 2x^2y^2 = 2a^2x^2 - 2a^2y^2$

$y^4 + 2(x^2 + a^2)y^2+ x^2(x^2 - 2a^2) = 0$

$y^2 = \frac{-2(x^2 + a^2) \pm \sqrt{[2(x^2 + a^2)]^2 - 4x^2(x^2 - 2a^2)}}{2}$

We need to keep the "+" solution else y isn't real. So:
$y^2 = \frac{-2(x^2 + a^2) + \sqrt{16x^2a^2 + 4a^4}}{2}$

$y^2 = \frac{-2(x^2 + a^2) + 2a\sqrt{4x^2 + a^2}}{2}$

$y^2 = -(x^2 + a^2) + a\sqrt{4x^2 + a^2}$

Thus
$y = \pm \sqrt{-(x^2 + a^2) + a\sqrt{4x^2 + a^2}}$

The area you are after will be between the "+" and "-" curves, and you know by symmetry they cross on the x-axis so you should be able to find these fairly easily. The integral you will be doing will be of the form:
$\int_a^b \sqrt{-(x^2 + a^2) + a\sqrt{4x^2 + a^2}}~dx$
which isn't going to be at all pretty. You'll likely have to do a numerical estimate. (Unless Krizalid or someone else pulls a trick out of their hat.)

-Dan

3. If $a>0$ this is the Leminscate of Bernoulli.

Express this curve in polar coordinates for simplification.

This is Mine 69th Post!!!

4. Hellow, kezman!

Find the area bounded by the curve given by the equation:
. . $(x^2 + y^2)^2 \:= \:2a^2(x^2 - y^2)$
I would convert to polar coordinates . . .

Conversions: . $x \: =\: r\cos\theta,\quad y \:=\: r\sin\theta,\quad x^2+y^2 \:=\: r^2$

Then we have: . $\left(r^2\right)^2 \;=\;2a^2\left(r^2\cos^2\!\theta - r^2\sin^2\!\theta\right)\quad\Rightarrow\quad r^4\;=\;2a^2r^2\left(\cos^2\!\theta - \sin^2\!\theta\right)$

. . which simplifies to: . $r^2\;=\;2a^2\cos2\theta$

This looks like a large Infinite-sign: $\infty$

. . It is symmetric to both the "x-axis" and the "y-axis".

We can integrate from $0$ to $\frac{\pi}{4}$ and multiply by 4.

. . $Area \;=\;4 \times \frac{1}{2}\int^{\frac{\pi}{4}}_0 2a^2\cos2\theta\,d\theta \;=\;4a^2\int^{\frac{\pi}{4}}_0\cos2\theta\,d\thet a$

Got it?

5. Yes! excelent all the responses. Thank you very much for the help-