# Multivariable: Area of region bounded by a spiral equation

1. Oct 20, 2011

1. The problem statement, all variables and given/known data

This is an example taken from the textbook lesson and there's one part I don't understand:

Find the area of the region bounded above by the spiral
r = pi/(3θ)

and below by the polar axis, between r = 1 and r = 2.

SOLUTION:

Double integral of r(dθ)(dr) with boundaries: r:[1,2] and θ:[0,pi/(3r)]

I don't understand why the inside integral (dθ) is taken from 0 -> pi/(3r) instead of just 0->pi/3 which is what I thought it should be.

2. Oct 20, 2011

Is it because one of the bounds must be a function of theta? And because the bounds of r are constants, then the bounds of theta must be of a variable? But this doesn't make sense to me because in a rectangular coordinate system, when you find the area of a rectangle using double integrals, both of the integrals have bounds of constants.

3. Oct 20, 2011

### LCKurtz

Have you drawn a graph? The easiest way to set this up is to break it into two integrals. The problem is that while the inner curve is r = 1, which curve you are on for the outer curve depends on θ. For θ from 0 to π/6 the inner and outer curves are the two circles. But from π/6 to π/3 the outer curve is the spiral.

4. Oct 20, 2011

Thank you very much! I appreciate your help, I actually grasped that earlier and realized that to be the problem (the book does give a picture and it draws dashed lines for pi/6 and pi/3 but says nothing about them). However, now I am having trouble grasping the concept as to how this integral even works... because with the bounds of r being designated to be from 1->2 (constants), how does using the variable equation pi/3r as the theta bounds end up accounting for the area region between pi/6 and pi/3 (by that I mean how does it end up subtracting this unwanted region)?

5. Oct 20, 2011

### LCKurtz

It is usually easier to set these types of problems up using the order rdrdθ instead of rdθdr for exactly the reason that is confusing you. That is because it is more natural to think of r as a function of θ than for θ as a function of r. If you must do θ first, you would go from θ on one boundary to θ on the other. The lower boundary is θ = π/6 and the upper boundary is θ on the curve, which is π/(3r), then r would go from 1 to 2.

It is more natural to have r go from 1 to π/3θ then θ go from π/6 to π/3.

Of course, this applies to just the top part of the area.

6. Oct 20, 2011

You said the lower boundary is θ = π/6 if you do θ first, do you mean to say the lower boundary is θ = 0?

7. Oct 20, 2011

### LCKurtz

No. Note the highlighted above. You do the problem in two pieces either way.

8. Oct 20, 2011

I apologize for this, I know you have thoroughly answered me but the book solves this problem with one double integral equation, that is (sorry I don't know how to use coding),

the double integral of rdθdr, with inner theta bounds 0->pi/(3r) and outer r bounds 1->2

to arrive at the final solution. Thoughts?

9. Oct 20, 2011

### LCKurtz

That is incorrect for the region described. It has to be done in two pieces because r on the outer curve is a two piece formula. You have drawn the graph, haven't you?

Hope that settles it because I have to leave for now. Good luck.