1
$\begingroup$

Problem: Let $a, b$ be positive real constants. Calculate the area of the set $$\mathcal{E} = \left\{ (x,y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\}$$

I am able to approach the problem with logical thinking like that:

For $a=b=1$ the set describes an unit circle. We can write the condition as $$ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 $$ which makes clear that the constants $a,b$ stretch the circle in $x,y$ direction. Thus we have an ellipse with the axis $a$ and $b$. The equation for the area of an ellipse is know to be $A = \pi a b$.

However, I guess that I am supposed to apprach the problem with some more general method (e.g. integrals, etc.). So how would I solve this problem in a more formal way? E.g. for sets with conditions like that in general?

$\endgroup$

3 Answers 3

2
$\begingroup$

The area that you want to compute is$$2b\int_{-a}^a\sqrt{1-\frac{x^2}{a^2}}\,\mathrm dx.$$Doing the substitution $x=at$ and $\mathrm dx=a\,\mathrm dt$, you get the integral$$2ab\int_{-1}^1\sqrt{1-t^2}\,\mathrm dt.$$But a primitive of $\sqrt{1-t^2}$ is $\frac12\left(t\sqrt{1-t^2}+\arcsin t\right)$. So,$$\int_{-1}^1\sqrt{1-t^2}\,\mathrm dt=\frac12\left(\arcsin(1)-\arcsin(-1)\right)=\frac\pi2$$and therefore your area is $\pi ab$, as you guessed.

$\endgroup$
10
  • $\begingroup$ Ok thanks, but how did you get from the problem to the first integral? I would be glad if you could explain it in general, like what would you do if there was a different equation? $\endgroup$ Commented Oct 26, 2017 at 10:19
  • $\begingroup$ @SimonMueller From $\frac{x^2}{a^2}+\frac{y^2}{b^2}\leqslant1$, I got$$-b\sqrt{1-\frac{x^2}{a^2}}\leqslant y\leqslant b\sqrt{1-\frac{x^2}{a^2}}.$$Therefore, that area that you are interested in is the area between the graphs of the functions $\pm b\sqrt{1-\frac{x^2}{a^2}}$, which is$$2b\int_{-a}^a\sqrt{1-\frac{x^2}{a^2}}.$$ $\endgroup$ Commented Oct 26, 2017 at 10:22
  • $\begingroup$ Thanks. And how would you proceed if you had a condition similar to that in general, like: $\{ (x,y)\in \mathbb{R}^2 : f(x,y) \leq c \}$? $\endgroup$ Commented Oct 26, 2017 at 10:29
  • $\begingroup$ @SimonMueller I am not aware of any general way of dealing with that. $\endgroup$ Commented Oct 26, 2017 at 10:30
  • $\begingroup$ Ok, one last question: Can you explain a bit further how you got from the equation to your first integral? This seems like a big jump to me. $\endgroup$ Commented Oct 26, 2017 at 15:34
0
$\begingroup$

The area is $$ \iint\limits_{\mathcal{E}}dx\,dy $$ With the substitution $x=a\rho\cos\varphi$, $y=b\rho\sin\varphi$ (with $\rho\ge0$ and $0\le\varphi<2\pi$), the limitation on $x$ and $y$ for $(x,y)\in\mathcal{E}$ become $$ \frac{a^2\rho^2\cos^2\varphi}{a^2}+\frac{b^2\rho^2\sin^2\varphi}{b^2}\le 1 $$ that is, $\rho^2\le1$ and so $0\le\rho\le1$; no further limitation on $\varphi$ is implied.

Why that substitution? Because we do know the set $\mathcal{E}$ is an ellipse, don't we?

The Jacobian is $ab\rho$, thus the integral becomes $$ \iint\limits_{\substack{\rho\in[0,1]\\\varphi\in[0,2\pi]}} ab\rho\,d\rho\,d\varphi = ab\int_0^{2\pi}\biggl(\int_0^1 \rho\,d\rho\biggr)\,d\varphi=\pi ab $$

$\endgroup$
6
  • $\begingroup$ Thanks. So to perform these steps you have to know that the equation describes an ellipse, correct? $\endgroup$ Commented Oct 26, 2017 at 15:31
  • $\begingroup$ @SimonMueller Not necessarily, but of course it helps. $\endgroup$
    – egreg
    Commented Oct 26, 2017 at 16:13
  • $\begingroup$ Ok, but you got to your substitution because you knew that it was an ellipse, correct? $\endgroup$ Commented Oct 26, 2017 at 21:07
  • $\begingroup$ @SimonMueller Well , yes, but it’s not so important: you just have to normalize the sine and cosine to use the fundamental identity. $\endgroup$
    – egreg
    Commented Oct 26, 2017 at 21:09
  • $\begingroup$ THanks, what do you mean by normalize? $\endgroup$ Commented Oct 26, 2017 at 21:11
0
$\begingroup$

Parametrize: $x=a \cos(\theta)$, $ y= b\sin(\theta),$ $0\le \theta \lt 2π.$

1st quadrant:

$\int_{0}^{a} ydx =$

$\int_{π/2}^{0} b\sin(\theta)(-a\sin(\theta))d\theta=$

$-ab\int_{π/2}^{0}\sin^2(\theta)d\theta=$

$ab\int_{0}^{π/2}\sin^2(\theta)d\theta.$

Note:

$ \int_{0}^{π/2}\sin^2(\theta)d\theta =$

$\int_{0}^{π/2}\cos^2(\theta)d\theta.$

Also:

$\sin^2(\theta)+\cos^2(\theta)=1.$

Now integrating LHS and RHS from $0$ to $π/2:$

$\rightarrow:$

$\int_{0}^{π/2}\sin^2(\theta)d\theta = π/4.$

Finally:

Area of the ellipse : $4 (π/4)ab =πab.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .