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 $\pi ab$

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    $\begingroup$ Maybe you can show us what you've done so we can give you suggestions on how to do the integral correctly $\endgroup$ – Alexander Vlasev Nov 2 '13 at 10:54

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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 $\pi ab$.

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    $\begingroup$ 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. $\endgroup$ – Vim Jun 23 '15 at 1:08
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    $\begingroup$ 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 $\int \frac{a}{b}f(x)dx = \frac{a}{b}\int f(x)dx$. $\endgroup$ – TonyK Aug 13 '16 at 13:54
  • $\begingroup$ @TonyK now this explanation should suffice for a rigorous one. $\endgroup$ – Vim Aug 13 '16 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: $$\pi ab$$

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    $\begingroup$ +1 for effort! MathJax tip: \sin u and \cos u look much nicer than sinu and cosu. $\endgroup$ – TonyK Aug 13 '16 at 14:05
  • $\begingroup$ Your 3rd to last line should be u+.5sin2u with the rest of the equation $\endgroup$ – Jinzu Jan 4 '18 at 18:27
  • $\begingroup$ @Jinzu I can't see where you got 5 from . $\endgroup$ – Kantura Jan 14 '18 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.

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    $\begingroup$ A warning for the reader: the second method is so severely mistaken that it might cause you brain damage. Better ignore it. $\endgroup$ – Alex M. May 11 '17 at 21:48

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

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

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


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