how do i find the area of the ellipse 4x^2 + y^2 = 9

using some form of integration by parts?

trig substitution?

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- Jan 14th 2009, 07:40 PMrazorfevertrig substitution and integration by parts
how do i find the area of the ellipse 4x^2 + y^2 = 9

using some form of integration by parts?

trig substitution? - Jan 14th 2009, 07:48 PMJhevon
- Jan 14th 2009, 07:52 PMrazorfever
yeah ... its an ellipse ... what was i thinking ... and i meant to say area of the circle ... i'm just used to finding the area under the circle ... wait i think u just answered my question ... would i have to integrate from 0 to the radius of the circle and multiply by four ??? but i still don't understand how to use trig substitution ... cause thats what i've been taught and i'm sure it fits in this question somewhere ...

- Jan 14th 2009, 07:54 PMpberardi
If you put this equation in terms of y you get y = sqrt(9-4x^2)

To get the interval of this equation find out where y = 0, the answer is +- 2/3

So the integral is [sqrt(9-4x^2) dx from -2/3 to 2/3

Use 9/4sin(theta) as a substitution for x and then integrate. - Jan 14th 2009, 07:55 PMMathstud28
You can, to do this solve for $\displaystyle y$ and notice that you will get a $\displaystyle \pm$...you can disregard this by doubling the area you get with $\displaystyle +$ (why?). You can then make a trigonometric sub of the form $\displaystyle 2x=3\sin(\vartheta)$

Alternatively we can parametrize this by $\displaystyle x=\frac{3}{2}\cos(\vartheta)$ and $\displaystyle y=3\sin(\vartheta)$. So now we can use the parametrization to find that the area is equal to $\displaystyle \frac{9}{2}\int_0^{2\pi}\cos^2(\vartheta)~d\varthe ta=\frac{9\pi}{2}$

Why does this work? Well if you know parametrization of curves find out that the area given by a counterclockwise oriented curve is $\displaystyle \int_{\alpha}^{\beta}x(t)\cdot y'(t)~dt$ where $\displaystyle [\alpha,\beta]$ is the interval on which the curve takes to make a full tracing. - Jan 14th 2009, 07:57 PMMathstud28
- Jan 14th 2009, 08:00 PMJameson
This isn't a circle. But yes you can use trig substitution. You have $\displaystyle y= \pm \sqrt{3^2-(2x)^2}$ Let the hypotenuse of a right triangle be 3, the vertical leg be 2x, and the other leg be 9-4x^2.

edit: well i am very far behind - Jan 14th 2009, 08:01 PMpberardi