I am supposed to determine the area of an ellipse using $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1.
I start by setting up an integral for the first quadrant, so I solved the above equation for y and got the integral $\int b\sqrt{1-\frac{x^2}{a^2}} $
The next step is to do a change of variables using u = $\frac xa$, and this is where I get a little confused. I then have to finish up by using geometry to evaluate the integral. Could someone explain how to do this? Thanks for any help in advance!
On
By substituting $u = \frac xa$ we get $$ \int_0^a b\sqrt{1-\frac{x^2}{a^2}}\,dx = ab\int_0^1 \sqrt{1-u^2}\,du.$$
Now substitute $u = \sin t$ to get
$$ \int_0^1 \sqrt{1-u^2}\,du = \int_0^{\pi/2} \cos t\sqrt{1-\sin^2 t}\,dt = \int_0^{\pi/2} \cos^2 t\,dt = \int_0^{\pi/2} \frac{1+\cos 2t}2\,dt.$$
Can you finish from here?
I think I understand what you want, but if I am mistaken please tell me.
We currently have the integral $$\int_0^a b\sqrt{1-\frac{x^2}{a^2}}dx$$ Make the substitution $u=\frac{x}{a}$. Hence $dx=a~du$. When $x=0$, $u=0$, and when $x=a$, $u=1$. Applying this to our integral we are left with $$\int_0^1 ab\sqrt{1-u^2}~du=ab\int_0^1 \sqrt{1-u^2}~du$$ Let's focus on the expression $$\int_0^1 \sqrt{1-u^2}~du$$ Using 'geometry', we can see that this is the area of a quarter-circle with radius $1$, ie it is equal to $\frac{\pi}{4}$. Hence one quarter of the area of an ellipse has area $$\frac{\pi ab}{4}$$ so the total area of the ellipse is, by symmetry, equal to $$\pi ab$$ Does that help? Please tell me if you have any questions.