# Double integration

1. Nov 15, 2005

Hi can someone help me with the double integration:
Sorry I dont know how to use the LATEX code.
it is a double integration
The first integration goes from 1/2 to1. the second integration goes from 0 to sqrt(1-x^2). there is no function after ,just dy dx.
I am asked to calculate the integral using polar coordinates.
I dont know how to find the bounds of integration corresponding for r.
Thank you
B

2. Nov 15, 2005

### quasar987

I love giving a shot at the problems you post Brad, they're always a tad more challenging than the calculus problems we usually see.

To convert the bounds to polar coordinates, you must use (fiddle with) the coordinate transformations equations: they are

x=rcosO
y=rsinO

and also the inverse ones (what you probably missed):

r²=x²+y²
O = Arctg(y/x)

You can also draw the surface of integration and try to figure out what the bounds are in terms of r and O.

3. Nov 16, 2005

### HallsofIvy

Staff Emeritus
First rule for any problem like this: draw a picture.
Since the "outer integral" is $\int_0^{\frac{1}{2}} dx$, x can run from 0 to $\frac{1}{2}$. The "inner integral" is $\int_0^\sqrt{1-x^2} dy$ so y ranges from 0 to $\sqrt{1-x^2}$.
$y= \sqrt{1-x^2}$ is the upper half of the circle $x^2+ y^2= 1$, the circle of radius 1 centered at (0,0). Draw that circle and the line x= 1/2 to see the area of integration.
Now, imagine an line from the origin "sweeping" over that area. The line starts horizontal and rotates up until it passes through $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. That is, the angle clearly starts at $\theta= 0[/tex] and goes to [itex]\theta= \frac{\pi}{3}$.
Radially, for each $\theta$ we move from the line x= 1/2 to the circle. $x= rcos(\theta)= \frac{1}{2}$ is the same as $r= \frac{1}{2cos(\theta)}= \frac{1}{2}sec(\theta)$. The limits of integration for the r integral are $\frac{1}{2}sec(\theta)$ and 1.
Don't forget to replace dxdy with $rdrd\theta$

Last edited: Nov 16, 2005
4. Nov 16, 2005

### quasar987

HallsofIvy posted the other way of doing it; what one might call the "geometric way", as opposed to the "algebraic way". If you doubt the answer you obtained through any of the two ways, do it the other way to see if you get the same thing.

5. Nov 18, 2005