Area bounded with curves I need to calculate area bounded with curves $(x^2+y^2)^2\le 3(x^2-y^2), x^2+y^2\le \sqrt{2}x$. Here we have inside of leminscate and circle. I think i should use that $P=\iint dxdx$ but not sure how.
 A: See the picture below:

Expressing the upper half of the region that you mentioned in polar coordinates, we get:

*

*$r^4\geqslant3r^2\cos(2\theta)\bigl(\iff r^2\geqslant3\cos(2\theta)\bigr)$;

*$r^2\leqslant\sqrt2\,r\cos(\theta)\bigl(\iff r\leqslant\sqrt2\cos(\theta)\iff r^2\leqslant2\cos^2(\theta)\bigr)$
We have\begin{align}3\cos(2\theta)\geqslant2\cos^2(\theta)&\iff6\cos^2(\theta)-3\geqslant2\cos^2(\theta)\\&\iff\cos(\theta)\geqslant\frac{\sqrt3}2\\&\iff\theta\geqslant\frac\pi6.\end{align}On the other hand, if $\theta\in\left[\frac\pi4,\frac\pi2\right]$, then $3\cos(2\theta)\leqslant0$. So, compute$$\int_{\pi/6}^{\pi/4}\int_{\sqrt{3\cos(2\theta)}}^{\sqrt2\cos(\theta)}\rho\,\mathrm d\rho\,\mathrm d\theta+\int_{\pi/4}^{\pi/2}\int_0^{\sqrt2\cos(\theta)}\rho\,\mathrm d\rho\,\mathrm d\theta.$$You should get$$\frac\pi6-\frac{3-\sqrt3}4$$and therefore the area that you're after is$$\frac\pi3-\frac{3-\sqrt3}2\approx0.413223.$$
A: Here are the first couple of steps to illustrate how to proceed.
Step: 1
Your domain of integration should satisfy both inequalities, i.e. you must integrate over the area where both
$$(x^2+y^2)^2 \le 3(x^2-y^2)$$
and
$$x^2+y^2 \le \sqrt{2}x$$
hold. If you plot the curves you'll obtain the following:

Therefore the area to be found is the one that is comprised in both curves, i.e. the area inside the lemniscate in the interval comprised between the origin and the left vertical dashed line plus the area inside the circle delimited by the two vertical dashed lines.
How do we know this is the right integration area?
We test it with points we choose "smartly". First, you plot the equations as in the image below, and then you test certain points to see if they satisfy both your domain conditions (i.e. the inequalities you start with). Since you have 2 inequalities, you have to test (in principle) 2*2 regions. If you look at the picture below, the points sketched are in taken in certain regions delimited by the two curves (the red points are symmetric, so you can test arbitrarily just one of them):
\begin{cases}
p_\text{red} = (1/2,\pm 1/2) \\
p_\text{green} = (1, 0) \\
p_\text{yellow} = (3/2, 0) \\
p_\text{blue} = (2, 0)
\end{cases}
Of course the smart choice is to choose the test points on one of the axes whenever possible, to simplify the calculations as much as possible, but also a random point would do the trick. If you choose the point $p_\text{green}=(1,0)$, you will realise that it is the only one that satisfies both inequalities, and therefore the area mentioned in the previous point is indeed the one we need to compute.

Step: 2
You now need to find the intersection points between the curves, so you know your extrema of integration. For instance, you integrate under the lemniscate between $(0,0)$ and the intersection point, while you integrate under the circle from the latter point until the point where the circle intersects the axis of the abscissae.

How to do it? Simply solve the system
\begin{cases}
(x^2+y^2)^2 = 3(x^2-y^2) \\[2ex]
x^2+y^2 = \sqrt{2}x
\end{cases}
In your case you will get 3 solutions: the trivial $x=0, y=0$ and the two points $p_1 = (x_p, y_p)$ and $p_2 = (x_p, -y_p)$. (I won't provide you with the solution though, so you have to work for it). Another point needed is the intersection between the circle and the axis of the abscissae, let's call it $p_3 = (\sqrt{2},0)$, which is trivial to compute.
You can now observe that the region you want to integrate is symmetric with respect to the axis of the abscissae, so you can limit yourself to find the surface between the lemniscate and the $y=0$ axis in the domain $x\in [0,x_p]$, sum it to the surface between the circle and the $y=0$ axis in the domain $x\in [x_p,\sqrt{2}]$, and finally multiply by 2 (to add the volume below the abscissae axis).
Step: 3
Now that you know your extrema of integration, what do you integrate? Well, you need to find the area between the lemniscate and the $y=0$ axis, therefore you must work on the equation of the lemniscate and transform it into the form $y=f(x)$. Same goes for the circle, that has to be transformed into a function of the form $y=g(x)$. When you perform this step be careful to consider the correct branch, since both functions have multiple values!
Step: 4
When all these steps are in place, your area will be given by:
$$P = 2\,\biggl[ \int_{0}^{x_p} f(x)dx + \int_{x_p}^{\sqrt{2}} g(x)dx\biggr]$$
Hint:
This is not the only way to approach this problem. You could very well transition into polar coordinates instead, given the high level of symmetry of your problem. It is up to you to use the method you are more comfortable with.
I hope this helped!
