Homework Help: Separable Differential Equation

1. Feb 11, 2010

crm07149

1. The problem statement, all variables and given/known data
The volume V of water in a particular container is related to the depth h of the water by the equation dV/dh below. If V = 0 when h = 0, find V when h = 4.

2. Relevant equations
dV/dh = 16 sqrt(4-(h-2)2)

V(0) = 0
V(4) = ?

3. The attempt at a solution

I have not gotten the ODE in a form that I can integrate using separation of variables. I have gotten to the point dV/dh = 16 sqrt(-h2+4h) but this doesn't help much.

Last edited: Feb 11, 2010
2. Feb 11, 2010

jks

Unless you miswrote the equation, you don't need to use separation of variables in this problem, since the right side is an equation of 1 variable. But you can do it like this:

dV/dh = √(4-(h-2)²)

move the dh over
∫dV = ∫√(4-(h-2)²) dh

V = (integrate that expression with respect to h) + C

3. Feb 11, 2010

crm07149

The problem was from my textbook under the "separable equations" section so I listed it as that.

Integrating the expression √(4-(h-2)²) is what I was having trouble with. I tried u substitution, integration by parts, and other methods for awhile. Am I missing something obvious to integrate the expression?

4. Feb 11, 2010

Staff: Mentor

I would try a trig substitution. To make things a little simpler, I would make an ordinary substitution u = h - 2, and then the trig substitution.

I usually draw a right triangle, with t being one of the acute angles, hypotenuse 2 and opposite (to angle t) side u. So the substitution here is sin t = u/2.

Don't forget that you'll need to undo two substitutions when you finally evaluate your antiderivative.