Find figure area using integral

Question

Find area of figure plot, which is formed by following equations: $$x^2+4y^2=8a^2$$ $$x^2-3y^2=a^2 $$ $$0 \leq a \leq x$$

I've started my solution with the following idea: it is easier to calculate mentioned figure area by separating it into 2 pieces, divided by vertical line going through the point of function plot's intersection. However, even considering that amount of calculations needed is really enormous. Does anyone have any ideas how to solve this task in a more convenient way?

Orange circled section area is the one needed (formed by two function plots intersection)

Answer 1

It is convenient to integrate the area along the $y$-direction, which is

$$\int_{-a}^a \left( \sqrt{8a^2-y^2} - \sqrt{a^2+3y^2} \right)dy$$ $$=4a^2 \int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \sqrt{1-t^2}dt - \frac{a^2}{\sqrt3}\int_{-\sqrt3}^{\sqrt3} \sqrt{1+t^2}dt$$ $$=4a^2 \cdot \frac{2+\pi}4 - \frac{a^2}{\sqrt3}\cdot (2\sqrt3 + \ln(2+\sqrt3))$$ $$=a^2 \left(\pi - \frac1{\sqrt3} \ln(2+\sqrt3)\right)$$

where the substitutions $t=\frac y{\sqrt2 a}$ and $t=\frac {\sqrt3 y}{ a}$ are used in the first and second integrals, respectively.

Answer 2

Some very practical hints, but not a complete solution

Replace $x' = ax, y' = ax$; the corresponding equations in $x', y'$ then do not involve $a$. If you find the area $U$ there, then the area in $xy$-coordinates will be $U/a^2$.

Once you've done that, you have (omitting the primes): $$ x^2 + 4y^2 = 8\\ x^2 - 3y^2 = 1 $$ It'd be a lot easier if the first equation didn't have different coefficients on $x$ and $y$, so substitute $x = 2x'$; then you have $$ 4x'^2 + 4y^2 = 8\\ 4x'^2 - 3y^2 = 1 $$ Again omitting primes, you get $$ x^2 + y^2 = 2\\ 4x^2 - 3y^2 = 1 $$ When you find the area in this last problem, you'll need to scale up by a factor of 2 to get the area in the 2nd-to-last problem.

Answer 3

HINT.-Your area is given by $$\dfrac{2}{\sqrt3}\int_a^{2a}\sqrt{x^2-a^2}dx+2\int_{2a}^{2\sqrt2a}\sqrt{8a^2-x^2}dx$$ whose bounds of integration come from $x^2=a^2$ and $x^2=8a^2$ with $a$ non-negative.

Both integrals are of elementary resolution.