Looking for HSC notes and resources? Check out our Notes & Resources page

Ellipse area (1 Viewer)

Thread starter Sunyata
Start date Mar 18, 2010

S

Sunyata

New Member

#1

Hi

Could someone please show me how to do this question?

Prove that the area of the ellipse x^2/a^2 + y^2/b^2 = pi*ab u^2

N

nonsenseTM

Member

#2

do you mean to prove the area of x^2/a^2 + y^2/b^2 = 1 is piab ?

N

nonsenseTM

Member

#3

sorry I don't konw how to integrate that

D

Drongoski

Well-Known Member

#4

$\100dpi From\ the\ eqn\ of\ the\ ellipse\ you\ get:\\ \\ y^{2}\ =\ \frac{b^{2}}{a^{2}} (a^{2}-x^{2})\ \ \ \Rightarrow \ \ \ y\ =\ \pm \frac{b}{a}\sqrt{a^{2}-x^{2}}\\ \\ Taking \ upper\ branch:\ \ y\ =\ \frac{b}{a}\sqrt{a^{2}-x^{2}}\\ \\ By\ symmetry\ area\ of\ ellipse\ =\ 4\times \frac{1}{4}\ area\ of\ ellipse\\ \\ =\ \ 4\int_{0}^{a}\frac{b}{a}\sqrt{a^{2}-x^{2}}\ dx\ \ =\ \ \frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}\ dx\\ \\ =\ \ \frac{4b}{a}\ \times\ \frac{1}{4}\pi a^{2} \ \ \ \ \ (integral\ =\ \frac{1}{4}\ area\ of\ circle\ radius\ a )\\ \\ =\ \ \pi ab\ \ u^{2}$

Last edited: Mar 18, 2010

You must log in or register to reply here.

Share:

Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Link

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top