# trig substitution and integration by parts

• Jan 14th 2009, 07:40 PM
razorfever
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?
• Jan 14th 2009, 07:48 PM
Jhevon
Quote:

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?
• Jan 14th 2009, 07:52 PM
razorfever
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 PM
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.
• Jan 14th 2009, 07:55 PM
Mathstud28
Quote:

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 $\displaystyle y$ 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 PM
Mathstud28
Quote:

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 $\displaystyle x=\pm\frac{3}{2}$
• Jan 14th 2009, 08:00 PM
Jameson
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 PM
pberardi
Quote:

Originally Posted by Mathstud28
Right idea, but be careful with your solving of this...it should be $\displaystyle x=\pm\frac{3}{2}$

ack! Just for the record I have +-3/2 written in my notebook (Nerd). I welcome the corrections.