how do i find the area of the ellipse 4x^2 + y^2 = 9
using some form of integration by parts?
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 ...
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.
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.