Integration changing to polar coordinates

[quietrain]

Homework Statement

[PLAIN][URL]http://i1115.photobucket.com/albums/k554/shirozack/polarchange.jpg[/PLAIN][/URL]

The Attempt at a Solution

why is the limits for the polar angle pi/3 to pi/2?

shouldn't it be pi/2 to pi/3?



because

x goes from 0 to 1/2, since x =rcos(T)
0 = cos(T) , T = pi/2

1/2 = cos(T), T = pi/3

so shouldn't the limits be from pi/2 to pi/3?

thanks!

 

[Sethric]

When you are changing coordinate systems with an integral, you don't change each integral bound independently (using the bounds on the x coordinate to find theta bounds, per say). You need to find the shape of whatever you are integrating and find new definitive bounds in your new coordinate system. In this case, your shape is a section of a circle. Find the bounds of that shape. Draw it if it helps.

 

[gb7nash]

Like Sethric said, get an idea of the shape before attempting the problem. For these types of problems, I always draw the graphs and shade in the appropriate region. From there, you can determine the start and finish angle.

 

[quietrain]

what if i can't draw out the region?

surely there must be a fix set of transformation rules right?

i mean, if i let x = rcos(T), then shouldn't the bounds automatically fall into place?

issn't this like substitution? where you solve for du = dx and correspondingly solve for the limits? i don't have to picture the graph right?

in any case,

the limits of y is actually a circle of radius 0 to 1 right?

but the limits of x is from 0 to 1/2. so can i assume that the angle is always from smaller to bigger? so it becomes pi/3 to pi/2? are there cases where i have to take a bigger angle to smaller angle?

 

[quietrain]

in fact, am i right to say that , if i integrate the polar angle, i should never get a negative right?

its just like integrating from 0 to 90, and 90 to 0, i will just get a minus sign for 1 of them right?

so i just take it away?

 

[Sethric]

Ok, um, a lot of questions.

If you can't even draw the shape, determining boundaries for the region become even harder. To my knowledge, there is no transformation rule that will make them fall into place. You do, however, have a set of equations that determine some boundaries for you. Those are the equations for the boundaries on the original integral. In this case:

$$y = \sqrt{3} x$$

and

$$y = \sqrt{1 - x^2}$$

If you did your polar substitutions, you would get:

$$r Sin(\theta) = \sqrt{3} Cos(\theta)$$

$$Sin(\theta) = \sqrt{3} Cos(\theta)$$

$$\theta = \frac{\pi}{3}$$

and (skipping some algebra)

$$r = 1$$

It still takes some vision to see how everything falls into place, but these can usually give you some boundary lines in your new coordinate system, if you couldn't draw the original.


This isn't like a single variable substitution, because, although x is a function of r and $$\theta$$, r and $$\theta$$ are both functions of x and y. My best advice is to just practice at drawing regions. Textbooks tend to use the same variety of boundaries, especially when you are expected to change coordinate systems.

the limits of y is actually a circle of radius 0 to 1 right?

The limits of r are from 0 to 1. In this case, one of the bounds of y was a semicircle of radius 1. It is an actual curve that serves as an upper bound. The lower bound for y was just a line with constant slope of root(3).

but the limits of x is from 0 to 1/2. so can i assume that the angle is always from smaller to bigger? so it becomes pi/3 to pi/2? are there cases where i have to take a bigger angle to smaller angle?

There are cases where textbooks will give you reversed boundaries, but it is rare. In general, use the orientation given in the problem to determine if you are going from smaller to bigger angle. Don't make the assumption that it will always be positive orientation. In this case, we used the lower bound of y to determine that the integral needed to start at $$\frac{\pi}{3}$$. Always use the information given. I would say, if you are in doubt, stick with positive orientation, but don't assume it is impossible to see otherwise.

 

[quietrain]

ah isee... thank you !