# Find figure area using integral

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) • To make it more clear added picture.. Mar 1, 2020 at 18:46
• What exactly does "square of figure" mean? Mar 1, 2020 at 20:21
• @KM101 its area, will fix it now Mar 1, 2020 at 20:22
• Oh, you were circling the region, got it. Mar 1, 2020 at 20:23
• (1) Put $x=a\sec t$ so you get using $\sec^2 t-\tan^2 t=1$$\int\sqrt{x^2-a^2}=\int a^2\tan^2 t\sec t dt=a^2\int-\sec tdt +a^2\int \sec^3 t dt$$ You do have $$\int\sec tdt=\ln|\tan t+\sec t|$$ Besides $$\int\sec^3 tdt=\dfrac 12\sec t\tan t+\dfrac12\int\sec tdt$$ You end with $$\dfrac{a^2}{2}(\sec t \cdot \tan t-\ln|\tan t+\sec t|$$ with$t=arc\sec\dfrac{x}{a}$Mar 1, 2020 at 21:37 ## 3 Answers 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. 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. • And why would area be$U/a^2$? Seems not that obvious to me.. Mar 1, 2020 at 20:30 • The primed coordinate system is basically the unprimed one divided by$a$. For$a = 3$, for instance, when$x = 1$, you have$x' = 3$. A$3 \times 3$box in the primed system has area nine; that's a$1 \times 1$box in the unprimed system, which has area one. Hence, divide by nine. (I was mistaken about the wrong-order thing a moment ago) Mar 1, 2020 at 20:32 • Thanks, got it. I've understood that unconsciously, but you've made that clear, will try out your tricks in a moment. Mar 1, 2020 at 20:34 • Frankly, I'd also swap$x$and$y\$, because I find it easier to integrate vertically...but that's just me. Mar 1, 2020 at 20:36

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.

• Yeah, that what I've meant under separating area in two pieces, however don't find these integrals that simple due to amount of calculations required to solve them.. Mar 1, 2020 at 20:57
• One of them with logarithms and the other with inverse trigonometric functions but quite known as elementary integrations. If you have difficulties for this i can give you the solution with intermediate calculations in comments. Mar 1, 2020 at 21:01
• if that is possible Mar 1, 2020 at 21:09
• Hm, it seems to me, that second integral must be divided by 2, not multiplied? Mar 2, 2020 at 11:08