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