# Find the Area of the ellipse

Given $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$

where $$a>0$$, $$b>0$$

I tried to make $$y$$ the subject from the equation of the ellipse and integrate from $$0$$ to $$a$$. Then multiply by $$4$$ since there are $$4$$ quadrants. $$Area=4\int^a_0\left(b^2-x^2\left(\frac{b^2}{a^2}\right)\right)^\frac{1}{2}dx$$

I can't get the answer $$\pi ab$$

• Maybe you can show us what you've done so we can give you suggestions on how to do the integral correctly Nov 2, 2013 at 10:54 In order to find the the area inside the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we can use the transformation $(x,y)\rightarrow(\frac{bx}{a},y)$ to change the ellipse into a circle. Since the lengths in the $x$-direction are changed by a factor $b/a$, and the lengths in the $y$-direction remain the same, the area is changed by a factor $b/a$. Thus $$\text{Area of circle} = \frac{b}{a}\times \text{Area of ellipse},$$

which gives the area of the ellipse as $(a/b\times\pi b^2)$, that is $\pi ab$.

• the area is changed by a factor b/a. Not to dispute your nice approach, that's more of what an engineer will go about solving this problem. In mathematics, a rigorous proof is preferable, though.
– Vim
Jun 23, 2015 at 1:08
• This approach looks fine to me. But to answer @Vim's objection, you can simply stretch the ellipse in the $y$-direction instead of squeezing it in the $x$-direction. Because everyone knows that $\int \frac{a}{b}f(x)dx = \frac{a}{b}\int f(x)dx$. Aug 13, 2016 at 13:54
• @TonyK now this explanation should suffice for a rigorous one.
– Vim
Aug 13, 2016 at 13:57

Here is my proof if it is any use to anyone.

The equation of an ellipse is given by: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Rearrange and write this ( the equation of an ellipse ) in terms of y: $$y=b\sqrt{1-\frac{x^2}{a^2}}$$ Lets find the area of one quarter of the ellipse and multiple that by 4 to get the area of the entire ellipse. We will integrate from $$x=0$$ to $$x=a$$. So we are looking for: $$4\int_0^a y.dx$$ Which is the same as ... $$eq1$$: $$4\int_0^a b\sqrt{1-\frac{x^2}{a^2}}.dx$$ Let: $$\sin u=\frac{x}{a}$$ Rearrange to get: $$x=a\sin u$$ Differentiate to get: $$\frac{dx}{du}=a\cos u$$ Therefore: $$dx=a\cos u.du$$ Sub this back into equation $$eq1$$. But we also need to find the range of $$u$$ values. Using $$\sin u = \frac{x}{a}$$ we can sub in our original limits of $$x=0$$ and $$x=a$$ to get the new limits in terms of $$u$$ of $$u=0$$ and $$u=\frac{\pi}{2}$$.

Subbing this back into $$eq1$$ as well we get $$eq2$$: $$4\int_0^\frac{\pi}{2} b\sqrt{1-(\sin u)^2}.a\cos u.du$$

From trigonometry we know that: $$\sin^2\theta + \cos^2\theta = 1$$ Therefore $$\sqrt{1-\sin^2u}.\cos u = \sqrt{\cos^2u}.\cos u = \cos^2u$$ Subbing this back into $$eq2$$ we get: $$4\int_0^\frac{\pi}{2} ab\cos^2u.du$$ Which is the same as: $$4ab\int_0^\frac{\pi}{2} \cos^2u.du$$ But again from trigonometry we know that: $$\cos^2\theta=\frac{1}{2}(1+\cos2\theta)$$ Subbing gives us: $$4ab\int_0^\frac{\pi}{2} \frac{1}{2}(1+\cos2u).du$$ Which is the same as: $$2ab\int_0^\frac{\pi}{2} (1+\cos2u).du$$ Now we can finally integrate to get: $$2ab[u+\frac{\sin2u}{2}]_0^\frac{\pi}{2}$$ Which goes to: $$2ab[(\frac{\pi}{2}+0)-(0+0)]$$ Which finally goes to: $$\pi ab$$

• +1 for effort! MathJax tip: \sin u and \cos u look much nicer than sinu and cosu. Aug 13, 2016 at 14:05
• Your 3rd to last line should be u+.5sin2u with the rest of the equation Jan 4, 2018 at 18:27
• @Jinzu I can't see where you got 5 from . Jan 14, 2018 at 9:04

Method 1 The area of a region $R$ in $2$D is given by $$A=\iint_R1\,dA.\tag{1}$$ Noting the symmetry in the $x$ and $y$ axes, the area in the first quadrant can be multiplied by 4. Rearranging the expression for positive y gives $$y = \frac{b}{a}\sqrt{a^2-x^2}.$$ The region can also be reduced to a single integral, so that (1) is equivalent to $$A = \frac{4b}{a}\int_0^a\sqrt{a^2-x^2}\,dx.$$ This can be evaluated using a trigonometric substitution, such as $x=a\sin t$. Calculating the corresponding limits, $x=0\Rightarrow t=0$ and $x=a\Rightarrow t=\pi/2$, and noting $dx=a\cos t\,dt$ gives the expression $$A = \frac{4b}{a}\int_0^{\pi/2}a^2\cos t\cdot\cos t\,dt=2ab\int_0^{\pi/2}\cos(2t)+1\,dt=2ab\cdot\frac{\pi}{2}=\pi ab.$$ Method 2 The area can also be calculated using a double integral, but this is much more difficult to evaluate. Parameterising the ellipse as $x(t)=a\cos t$ and $y(t)=b\sin t$, (1) can be written as $$A=\int_0^{2\pi}\int_0^{R(t)}r\,dr\,dt,$$ where $R(t)$ is the boundary of the ellipse, given by $R(t)=\sqrt{x^2+y^2}=\sqrt{a^2\cos^2t+b^2\sin^2t} = \frac{1}{2}\pi(a^2+b^2)$. I'll leave the details here for you to confirm that the result is the same as the other techniques.

• A warning for the reader: the second method is so severely mistaken that it might cause you brain damage. Better ignore it. May 11, 2017 at 21:48
• Thanks for the warning, Alex. I did actually get confused. The distinction parametric/polar in this pdf helped me: qc.edu.hk/math/Resource/AL/… Oct 27, 2019 at 10:04

You can also use Green Theorem:

first, parameterize the ellipse

$$\begin{cases} x = a\sin t\\ y = b\cos t \end{cases}$$

Then by using Green Theorem

$$\int\int_{region} 1 \,dA = \int_{boundary} x \,dy$$

$$\int_{0}^{2\pi} a\cos t \,d(b\sin t) = \int_{0}^{2\pi} ab (\cos t)^2 \,dt = ab\pi$$

from the equation of the ellipse: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ $$y = \sqrt{\frac{a^2b^2 - x^2b^2}{a^2}} = \frac{b}{a}\sqrt{a^2-x^2}$$

since proceding with change of variable in integration is, using x = a*cos(Θ):: $$\\ \sqrt{a^2 - x^2} = \sqrt{a^2-a^2cos^2\theta} =asin\theta \\ \int_{}^{}\sqrt{a^2 - x^2} = a\int_{}^{}sin\theta \ dx \ = a^2\int_{}^{}sin^2\theta d\theta = a^2\int_{}^{}\frac{1 - cos 2\theta}{2} d\theta = \frac{a^2\theta}{2}-\frac{a^2sin^22\theta}{2}+C = \frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x\sqrt{1-\frac{x^2}a^{}}}{a})+C$$ so the area in the first quadrant is:

$$\frac{b}{a} \int_{0}^{a} \sqrt{a^2-x^2} dx$$ that is: $$\frac{b}{a}((\frac{a^2}{2}(cos^{-1}\frac{x}{a}-\frac{x \sqrt{1-\frac{x^2}{a^2}}}{a}))|^a_0) = \frac{ab \pi}{4}$$

so for the area of the entire ellipse: $$\frac{ab\pi}{4} 4 = ab\pi$$