Question

Prove that the area of an ellipse is pi(ab) using integration.

Answer 1

The equation of the ellipse shown above may be written in the form
\[x ^{2}/a ^{2}+y ^{2}/b ^{2}=1\]

Answer 2

http://math.ucsd.edu/~wgarner/math10b/pdf/area_ellipse.pdf

Answer 3

thanks

Answer 4

Since the ellipse is symmetric with respect to the x and y axes, we can find the area of one quarter and multiply by 4 in order to obtain the total area.

Solve the above equation for y
\[y=+ or - b \sqrt{1-x ^{2}/a ^{2}}\]

The upper part of the ellipse (y positive) is given by
\[y=b \sqrt{1-b ^{2}/a ^{2}}\]

We now use integrals to find the area of the upper right quarter of the ellipse as follows

(1/4) Area of ellipse = \[\int\limits_{0}^{a}b \sqrt{1-b ^{2}/a ^{2}}dx\]

Answer 5

that pdf is the hard way :)

Answer 6

what's the easy way?

Answer 7

waiting to see what floros comes up with

Answer 8

i just went through the proof i posted have you done trig sub yet?

Answer 9

zarkon if not busy wanna try http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e55c6c40b8b9ebaa890810e

Answer 10

if you know the area of the unit circle then you don't need to use trig sub

Answer 11

We now make the substitution sin t = x / a so that dx = a cos t dt and the area is given by
(1 / 4) Area of ellipse = \[\int\limits_{0}^{\pi/2} ab \sqrt{[1-\sin ^{2}t]} \cos t dt\]

\[\sqrt{1-\sin ^{2}t} = \cos t \] since t varies from 0 to pi/2, hence

Answer 12

(1 / 4) Area of ellipse = \[\int\limits_{0}^{\pi/2} ab \cos ^{2}t dt\]


Use the trigonometric identity \[\cos ^{2} t= (\cos 2t+1) / 2\] to linearize the integrand;

Answer 13

\[A/4=\int b\sqrt{1-x ^{2}/a ^{2}}dx\]


let \[u=\frac{x}{a}\]
\[a\,du=dx\]

\[=\int ab\sqrt{1-u^2}dx\]
but
\[\int \sqrt{1-u^2}dx\]

is just the area of one quarter of the unit circle so the area is \[\pi/4\]

thus

\[A/4=\int b\sqrt{1-x ^{2}/a ^{2}}dx=ab\pi/4\]

so

\[A=ab\pi\]

Answer 14

(1 / 4) Area of ellipse = \[\int\limits_{0}^{\pi/2} ab (\cos 2t +1)/2 dt\]

Evaluate the integral
(1 / 4) Area of ellipse = (1/2) b a [ (1/2) sin 2t + t]\[_{0}^{\pi/2}\] = (1/4) pi ab

Obtain the total area of the ellipse by multiplying by 4
Area of ellipse = \[4*(1/4) \pi ab = \pi ab\]

Answer 15

Alright guys, I think all of your answers are good. I think I'm too sleepy to fully grasp it at the moment, but I'll look at it again tomorrow. That's for all your help!