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, 08:40 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, 08:48 PMJhevon
- Jan 14th 2009, 08:52 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, 08:54 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, 08:55 PMMathstud28
You can, to do this solve for and notice that you will get a ...you can disregard this by doubling the area you get with (why?). You can then make a trigonometric sub of the form

Alternatively we can parametrize this by and . So now we can use the parametrization to find that the area is equal to

Why does this work? Well if you know parametrization of curves find out that the area given by a counterclockwise oriented curve is where is the interval on which the curve takes to make a full tracing. - Jan 14th 2009, 08:57 PMMathstud28
- Jan 14th 2009, 09:00 PMJameson
This isn't a circle. But yes you can use trig substitution. You have 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, 09:01 PMpberardi