# Homework Help: Find the multiple integral for a modeled problem

1. Nov 16, 2016

### xipe

1. The problem statement, all variables and given/known data
A 12 foot light pole stands at the corner of an 8 foot by 10 foot rectangular picnic blanket spread out on the ground. A bee flies in a straight line from a point P on the pole to a point Q on the blanket.

Set up a multiple integral whose value represents the average length of all possible paths along which the bee could fly.

2. Relevant equations
None so far.

3. The attempt at a solution
After sketching the situation, I think spherical coordinates would be best to use, but I am having trouble figuring out the bounds. I think that rho would go from sqrt(12^2+8^2) to sqrt(12^2+10^2+8^2), these being the shortest and longest length of rho from the top of the pole to the corners of the blanket. This would only give half of the region of the blanket though. If i could find an integral and multiply it by 2, it should work. If this were the case, theta would go from 0 to pi/2. My main issue would be finding the values for phi. I would imagine it would go from (pi/2)+tan^-1(12/sqrt(10^2+8^2) to pi. My other concern is that maybe rho has to be a function of phi somehow. Sorry for the confusing description, but this problem is giving me a lot of issues. Any advice to point me in the right direction would be immensely appreciated! Thank you!

2. Nov 16, 2016

### BvU

Hi xipe,
why do you think that ? Suppose I would think Cartesian coordinates were the best, how would you weigh our mututal proposals ?

3. Nov 16, 2016

### PeroK

I can't see how spherical coordinates will help here. Quite the opposite!

4. Nov 16, 2016

### xipe

So if I used Cartesian coords, would it just be 0<=x<=8,0<=y<=10,0<=z<=slope of the hypotenuse from the top of the pole to out outskirts of the blanket? My difficulty is accounting for the shape of the area of integration.

5. Nov 16, 2016

### Ray Vickson

If the bee flies from (0,0,z) to (x,y,0) you do not need any restrictions on the slope of the hypotenuse. The only restrictions are 0 ≤ z ≤ 12, 0 ≤ x ≤ 8 and 0 ≤ y ≤ 10.

6. Nov 16, 2016

### xipe

That's what I thought originally, but wouldn't that give you a cubic volume when you integrate it? I'm missing something, otherwise the integrand should just be dxdydz?
And to find the average value, the volume of the region of integration is another confusing thing to figure out.

7. Nov 16, 2016

### BvU

Does that give you a length ?
Draw the line from (0,0,z) to (x,y,0) and write an expression for the length ...

8. Nov 16, 2016

### xipe

Would that length be sqrt(z^2+sqrt(x^2+y^2))?
I apologize for all of the questions and greatly appreciate all of your help.

9. Nov 16, 2016

### PeroK

You don't need that second square root. Just use the distance in Cartesian coordinates.

10. Nov 16, 2016

### Ray Vickson

The integration element is just $dx\, dy \, dz$, but the integration region is not a rectangular box; you are not looking at a solid region where a little $dx\, dy$ cylinder segment goes from $(x,y,0)$ to $(x,y,z_{\max}(x,y))$, and you most definitely are not evaluating the volume of a solid region.

If I were doing the problem I would do the $xy$-integration first, for fixed $z$, then integrate over $z$ after that.

Last edited: Nov 16, 2016