# Thread: trig substitution and integration by parts

1. ## trig 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?

2. Originally Posted by razorfever
how do i find the area under the circle 4x^2 + y^2 = 9
using some form of integration by parts?
trig substitution?
firstly, this is not a circle. secondly, what do you mean by "under"? below the curve but above the x-axis?

3. 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 ...

4. 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.

5. Originally Posted by razorfever
how do i find the area under the circle 4x^2 + y^2 = 9
using some form of integration by parts?
trig substitution?
You can, to do this solve for $y$ and notice that you will get a $\pm$...you can disregard this by doubling the area you get with $+$ (why?). You can then make a trigonometric sub of the form $2x=3\sin(\vartheta)$

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

Why does this work? Well if you know parametrization of curves find out that the area given by a counterclockwise oriented curve is $\int_{\alpha}^{\beta}x(t)\cdot y'(t)~dt$ where $[\alpha,\beta]$ is the interval on which the curve takes to make a full tracing.

6. Originally Posted by pberardi
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.
Right idea, but be careful with your solving of this...it should be $x=\pm\frac{3}{2}$

7. This isn't a circle. But yes you can use trig substitution. You have $y= \pm \sqrt{3^2-(2x)^2}$ 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

8. Originally Posted by Mathstud28
Right idea, but be careful with your solving of this...it should be $x=\pm\frac{3}{2}$
ack! Just for the record I have +-3/2 written in my notebook . I welcome the corrections.