Thread: Unusual use for integral, explanation why?

1. Unusual use for integral, explanation why?

Hi,

So I have the following problem:

A vessel is in the form of an inverted circular cone with a semi-vertical angle of 30 degrees. Water is poured in at 5cm^3/min and leaks out from a hole at the bottom at a rate of $\frac{\sqrt{h}}{30}$cm^3/min. Set up the integral that gives the time for the depth to change from 0.5cm to 5cm.

I have got the (correct) integral:

$\frac{dt}{dh} = \frac{10*\pi*h^2}{150-\sqrt{h}}$

Then the answers say I should take the integral of the above equation, with respect to h, from h=0.5 to h=5.

I'm wondering, why? Doesn't taking that integral just find the area of the curve? How does it relate to the question? Or does doing that subtract the time at t=5 from the time at t=0.5?

Just confused about the why. Thanks!

2. You are right! It does find the area UNDER the curve. However, it is not just any ordinary area. The x-axis (or h-axis) measures distance (that is length). The y-axix measures dt/dh which is the rate of change of time with respect to distance (h). That is the y axis measures time/length. If you have a square with corners (0,0),(a,0),(0,b),(a,b) in the dt/dh versus h plane, it's area is just a*b and the units are length*time/length = time. So the area actually signifies time!

3. Ah, much clearer now. Thanks heaps!