Calculate the area bounded by two curves via double integration

Question

I need to find the area of the middle part bounded (or between) 2 curves: $ x²+y²=1$ and $ 4x²-y²=1$.

I have the graphic of the middle part (the part, which I need to calculate the area for it), but I can't understand, do I need to solve this in polar system or Cartesian?

I think, that I can move to Cartesian, but I don't know which function to integrate.

In Cartesian system I'm solving this integral: $$\int_{-\frac{\sqrt{1+y^2}}{2}}^{\frac{\sqrt{1+y^2}}{2}}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}1\ dx\ dy$$

But I'm getting a function of x or y, not a number.

And in polar system I can't understand which function to use, I know that $\ r$ will vary from $\ 0$ to $\ 1$ and $\theta$ will vary from from $0$ to $\ 2\ π$.

Answer 1

Hint: The area of one of the unshaded parts is given by

$$\int_{-\sqrt{\tfrac35}}^{\sqrt{\tfrac35}} \int_{\tfrac{\sqrt{1+y^2}}2}^{\sqrt{1-y^2}} dx \, dy$$

which can be used to determine the desired area.

Answer 2

I can't see how to do this without breaking this integral in two.

In Cartesian, you can either integrate with respect to y first and integrate the caps separate from the area bounded by the parabola.

Or, with respect to y first and do the middle bounded by the circle separate from what is bounded at one end by the parabola and at the other by the circle.

In either case, you need to find the points where the curves intersect.

Or you can do this in polar

It will still be two integrals.

Where do the curves intersect?

$x^2 + y^2 = 1\\ 4x^2 - y^2 = 1\\ 5x^2 = 2\\ x = \sqrt {0.4}\\ y = \sqrt {0.6}$

$4\int_{0}^{\sqrt{0.6}}\int_0^{\frac {\sqrt {1+y^2}}{2}} \ dx \ dy + 4\int_{\sqrt{0.6}}^{1}\int_0^{\sqrt {1-y^2}} \ dx \ dy$

or
$4\int_0^\frac 12 \int_0^{\sqrt{1-x^2}}\ dy\ dx + 4\int_{\frac 12}^{\sqrt{0.4}} \int_{\frac {\sqrt {1+y^2}}2}^{\sqrt{1-y^2}} \ dx\ dy$

or, in polar.

$4\int_0^{\arctan \sqrt {\frac 32}}\int_0^{\frac {1}{4\cos^2\theta + \sin^2\theta}} r\ dr\ d\theta + 4\int_{\arctan \sqrt {\frac 32}}^{\frac {\pi}{2}}\int_0^1 r\ dr\ d\theta$