Water Pressure in Centrifuge

[bphys348]

Homework Statement

A test tube filled with water is being spun around in an ultra centrifuge with angular velocity ω. The test tube is lying along a radius, and the free surface of the water is at a radius r₀.

(a) Show that the pressure at radius r within the test tube is P = (1/2)rho*ω²(r²-r₀²). Ignore gravity and atmospheric pressure

Homework Equations

dP= -ρgdz

The Attempt at a Solution

Since the test tube is undergoing circular motion, and the water within the test the test tube lies at a varying radius from center, the acceleration (g) is given by a=(v²)/r = ω²r

So dP = -ρgdz = -ρω²rdr

P= -(1/2)ρω²r² from r to r₀

I think that much is fine, and this gives me the correct answer... But, here is where my understanding is hazy. If I define r to be increasing radially from center, then it is my understanding that my lower bound should be r and my upper to be r₀.

Here's my question: If I define an axis to be increasing in one direction, then should I always integrate from the lower value to higher?

If I use this same reasoning with a simple pressure vs. water depth example then I run into trouble. If I want to find the pressure at a depth of 30m and I define my z axis to be increasing upwards then it seems to me the integral should look like this:

dP = ∫-ρgdz from -30 to 0 = -ρg(0 - -30) = -ρg*30

But the answer should be positive!

Any help would be much appreciated

 

[Orodruin]

In the formula you have used, the gravitational acceleration is pointing in the negative z direction. If you change the direction of the axis, then you must also change the direction of the field.