Separable Differential Equation

[crm07149]

Homework Statement

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.

Homework Equations

dV/dh = 16 sqrt(4-(h-2)²)

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

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.

 

[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

 

[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?

 

[Mark44]

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.