# Integrating with polar co-ordinates

1. Jan 7, 2009

### franky2727

revising for an exam and have the question let I=∫lim ∞ to 0 e-x2 dx. since x is a dummy variable here, replace it with y to get a second expansion for I. Multiply these two together to get a double integral for I2. Transform into polar co-ordinates noting that dxdy corresponds to r dr d♂. Carry out the integrations and deduce I.

Right so i got the new equation and multiplied together to get r2cos2♂+r2sin2♂ giving me a lovely 1*r2 and giving e-r2

after which i get stuck because i dont understand the limits on the first integral♂ change from inf-0 to pi/2-0 answers on a postcard please!

2. Jan 7, 2009

### Dick

Ok, so you've gotten the integral to a double integral over the first quadrant of e^(-r^2)*dx*dy. Now you want to change to polar coordinates r and theta. As the instructions say, dx*dy becomes r*dr*dtheta. You also need to change the x and y integration limits to limits for r and theta. What are the limits on r and theta that will cover the first quadrant?

3. Jan 7, 2009

### franky2727

i guess pi/2 and 0 but only because i already have the answers, where do these first quadrant results come from? and why dont the y values change?

4. Jan 7, 2009

### Dick

You are integrating over the region of the (x,y) plane where x>=0 and y>=0. It's the first quadrant. You describe those point is polar coordinates by describing r, the distance from the origin, and theta, the angle the point makes with respect to the x axis. What range of r and theta will cover the first quadrant??

5. Jan 7, 2009

### franky2727

ah, zero rad when working up the y-axis positive part and 90degree of pi/2 when working across the positive x-axis, so because the limit is infinity and zero the area is the whole space so pi/2 and zero. Still think i'm a bit lost, like how do we know were working in the first quadrent ? (x>0 and y>0) and how would we change other limits such as pi and 0 or 2 and 1

6. Jan 7, 2009

### Dick

You know x>=0 and y>=0 because those are the limits on your two integrals for e^(-x^2) and e^(-y^2) in terms of x and y. You figure out theta limits in other cases by drawing a picture of the integration region and looking at what the appropriate angles are.

7. Jan 7, 2009

### franky2727

so both the integrals are changed together? i dont see how i can get an angle when i just have values of one variable, for instance given the x integral with limits 6 and 1 say, what does 6 and 1 show on the first quadrent graph? surely this is just the y=6 line and the y=1 line? no angles

8. Jan 7, 2009

### Dick

You are changing the product of two single integrals dx and dy into a SINGLE double integral dx*dy.

9. Jan 7, 2009

### Staff: Mentor

No, that's not it. Zero (radians) corresponds to the direction of the positive x-axis, and pi/2 (rad) is the direction of the positive y-axis.

For this question, the limits of integration, 0 to infinity, for both x and y, correspond to 0 <= x <= inf, 0 <= y <= inf. It should be obvious that this is the first quadrant.

If you had an iterated integral with these limits:
$$\int_{-\infty}^{\infty}\int_0^{\infty}{...}dy dx$$
the region over which you're integrating is the half plane where y >= 0. If you converted to polar form, the limits of integration would be 0 and infinity for r, and 0 and pi for theta.

10. Jan 7, 2009

### franky2727

ok so in this case we get integral of infinity squared and 0 or just infinity and zero but i still dont see how this gives us a line which we can use to get an angle from :S sorry i must be being really slow here, is there anyway you can show it graphically?

11. Jan 7, 2009

### franky2727

can you give a few examples using a mix of rational numbers, pi values and infinty/0 to see if i can figure out whats happening from those transformations?

12. Jan 7, 2009

### Staff: Mentor

Nope, there's no infinity squared going on.

I can't show it graphically using text, and I'd rather not take the time to draw something. Here's one for you to try:

$$\int_{-1}^{1}\int_0^{\sqrt{1 - x^2}}{...}dy dx$$

What I'd like you to do is to describe the region over which integration is to be performed.

Next, describe what the limits of integration would be in a polar iterated integral. If you understand what the integration region is, converting the limits of integration to polar form is pretty easy.

13. Jan 7, 2009

### franky2727

so by multiplying the two integrals? we will get between root 1-x^2 and zero? i'm sorry i really really dont even know where were supost to be starting here, are we ment to use the formula? r^2=x^2+y^2 and tantheata=y/x?

14. Jan 7, 2009

### franky2727

quoting you from a previous example on this

Here's why you have the bounds you show.
The region of integration for the iterated integral in rectangular coordinates was the first quadrant, 0 <= x <= infinity, 0 <= y <= infinity.
The region of integration won't change in switching to polar form, but its description usually does. Your first integration is with respect to r, so you want r to range from 0 to infinity. Your second integration is with respect to theta, so you want theta to range from 0 to pi/2. Your limits of integration are switched.

i'm with you until "Your first integration is with respect to r, so you want r to range from 0 to infinity" where did that come from, i realise were in the first quadrent and that all were doing is changing the variables so no size value is going to change just the measures we are using to describe it. ( a bit like changing from pounds to dollars) the thing i'm bemused about is where you are getting the exchange rate from

15. Jan 7, 2009

### Staff: Mentor

Right, the region of integration stays exactly the same, but how it's described (rectangular vs polar coordinates) changes.

Here's the example again:
$$\int_{-1}^{1}\int_0^{\sqrt{1 - x^2}}{...}dy dx$$

Describe the region over which integration is to be performed.

Next, describe what the limits of integration would be in a polar iterated integral. If you understand what the integration region is, converting the limits of integration to polar form is pretty easy.

16. Jan 8, 2009

### franky2727

the region is between the line y=root1-x^2 and y=0 for the y integral and between x=1 and x=-1 for the x integral

17. Jan 8, 2009

### franky2727

been looking through some more online examples but to no avail, think i'm getting a little bit closer tho, i beleive from the limts we get a "domain" such as this one i found x^2+y^2are smaller than or equal to 9 but bigger than or equal to 4 and yis bigger than or equal to zero, so from this i can see that its in the 4th and 1st quadrent, and that the lower limit of x or y is 2 and the upper is 3 so this gives me theata between 0 and pi from the quadrent information and r is between 3 and 2 from the numbers information, the question i will ask now is how to find this domain from the dy*dx limits?

18. Jan 8, 2009

### franky2727

i think if someone told me what exactly is being multiplies when multiplying dx*dy i might be able to get this now

19. Jan 8, 2009

### gabbagabbahey

$dxdy$ is an infinitesimal area element of width $dx$ and height $dy$.

In some cases, it is easier to use polar coordinates $\rho$ and $\phi$ (or r and theta). In that case, the infinitesimal area element is the small patch of area with radial component of $d\rho$ and arc length of $\rho d\phi$ since these are both infinitesimally small lengths, they are essentially perpendicular and so their product gives the area element (i.e. $da=\rho d\rho d\phi$ in polar coordinates)

Now, when you try to integrate

$$I^2=\int_0^{\infty}\int_0^{\infty}e^{-(x^2+y^2)}dxdy$$

The area that are integrating over is the infinite quarter plane that lies in the first quadrant. Is this region rectangular or circular? The answer is both! Picture a quarter circle in the first quadrant with radius $\rho$....what happens as $\rho\to \infty$ ? What area is enclosed? The answer is the infinite quarter plane in the first quadrant.

Now, what does $$e^{-(x^2+y^2)}$$ become in polar coordinates?

What does $da$ become in polar coordinates?

You know that the limits on $\rho$ must be from zero to infinity; what about the limits on $\phi$?

Last edited: Jan 8, 2009
20. Jan 8, 2009

### HallsofIvy

Staff Emeritus
And what geometric figure is $y= \sqrt{1- x^2}$? Not in terms of x and y- the geometry of the situation is what you want.