I have to find an area using definite integral, where the figure is restricted by two functions, stated in title. Can someone explain how to do that?
Here is my try:
So I equal the expressions to find intersections, - $$\frac{a^3}{a^2+x^2}=\frac{x^2}{2a}$$
$$\frac{2a^4}{a^2+x^2}=x^2$$
$$2a^4=a^2x^2+x^4$$ therefore $${2a^4-a^2x^2-x^4}=0$$
$$x=\frac{a^2 \pm \sqrt{a^4-4 \cdot (-1) \cdot 2a^4}}{{-2}}=\frac{a^2 \pm \sqrt{9a^4}}{-2}$$ $$x_1=\frac{4a^2}{-2}=-2a^2$$ and $$x_2=a^2$$
I don't know what to do next, and doubt that my calculations are correct.
When you are applying the quadratic equation, you are solving for $x^2$, not $x$ since this is a fourth degree polynomial equation which is quadratic in form.
The first solution is not possible because $x_1^2 $cannot be negative.
So that leaves the second solution $x_2^2=a^2$.
So you get two solutions for $x$.
You get $x=\pm a$.
So to get the area of your figure you must subtract the lesser of the two functions from the greater and integrate the difference over the interval $[−a,a]$.
So you must solve the integral
$$ \int_{-a}^a\frac{a^3}{a^2+x^2}-\frac{1}{2a} x^2\,dx$$
The first term will involve the arctangent function and the second term is a polynomial.
NOTE: Since the region is symmetric with respect to the $y$-axis, you can find the area by doubling the integral from $0$ to $a$, simplifying the process somewhat.