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 $\pi ab$
$\endgroup$ 15 Answers
$\begingroup$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 $\pi ab$.
$\endgroup$ 3 $\begingroup$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: $$\pi ab$$
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ 2 $\begingroup$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
$$ \int\int_{region} 1 \,dA = \int_{boundary} x \,dy $$
$$\int_{0}^{2\pi} a\cos t \,d(b\sin t) = \int_{0}^{2\pi} ab (\cos t)^2 \,dt = ab\pi$$
$\endgroup$ $\begingroup$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:$$ \frac{ab\pi}{4} 4 = ab\pi$$
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