# Intgral of dp^2?

1. Aug 25, 2011

### hermano

intgral of dp^2??

Hi,

I have the following equation:

$\dot{m} = c \frac{dp^{2}}{dx}$

Integration gives:

$\dot{m} \int dx = c \int dp^{2}$

How can I solve $\int dp^{2}$??

The lower boundary is Ps and the upper boundary is P0 for the integral!

2. Aug 25, 2011

### tiny-tim

hi hermano!

do you mean dq/dx where q = p2 ?

just integrate over q from q = (ps)2 to q = (p0)2

3. Aug 25, 2011

### hermano

Re: intgral of dp^2??

Hi tiny-tim ;-)

I want to integrate dp^2!! So you take some parameter q and say that this parameter q is equal to p^2 and then solve the integral of dq instead of dp^2. This gives q again and thus p^2! Then fill in the upper and lower boundary for p: P0^2 - Ps^2

Am I right?

4. Aug 25, 2011

### tiny-tim

if i'm correctly understanding your original equation, yes

5. Aug 25, 2011

### hermano

Re: intgral of dp^2??

My original question is how to solve this equation:

$\dot{m} \int dx = c \int dp^{2}$

My problem is solving the right hand side of the equation namely:

$c \int dp^{2}$

Your answer is: replace $p^{2}$ by q and solve this 'normal' integral $c \int dq$! This gives c q, then replace q by $p^{2}$ again so the solution is c $p^{2}$. Then I only have to fill in the boundaries for p. Right?

6. Aug 25, 2011

### tiny-tim

I think so, yes.

7. Aug 25, 2011

### K^2

Re: intgral of dp^2??

dp²/dx is an infinitesimal quantity. On the left side of equation, you have a finite quantity. Something is wrong.

You either have (dp/dx)² or d²p/dx². In either case, review your equation. Where do you get it from, anyways?

8. Aug 26, 2011

### hermano

Re: intgral of dp^2??

Hi K^2,

This equation comes from Darcy's law namely $\dot{m} = \frac{p Q}{R T}$

Fill Q (volumetric flow rate form Darcy's law) in in the above equation gives:

$\dot{m} = c p \frac{dp}{dx}$

With $c$ all the constants in the equation.

$p \frac{dp}{dx}$ is equal to $\frac{1}{2} \frac{dp^2}{dx}$

And thus finally I got $\dot{m} = \frac{c}{2} \frac{dp^2}{dx}$

Solving this:

$\dot{m} \int dx = \frac{c}{2} \int dp^{2}$

Therefore my question: How to solve $\int dp^{2}$ ??

9. Aug 26, 2011

### tiny-tim

ah, now i see where it comes from!

yes, the whole point of changing pdp/dx to 1/2 d(p2)/dx

(and btw, it's better to write it that way, with brackets)

is because that is a perfect derivative, and you can immediately integrate it to pf2 - pi2.

Incidentally, the same trick enables us to go from F = ma (in mechanics) via a = vdv/dx to ∆(1/2 mv2)

10. Aug 26, 2011

### hermano

Re: intgral of dp^2??

Tiny-tim,

Do I make a fault against the mathematics by saying that $\dot{m} = \frac{c}{2} \frac{dp^2}{dx}$ can be solved through $\dot{m} \int dx = \frac{c}{2} \int dp^{2}$??

Then still my question is: how to solve the right hand site of this equation namely the integral of $\int dp^{2}$ ??

11. Aug 26, 2011

### K^2

Re: intgral of dp^2??

Yeah, you shouldn't write it as dp²/dx, because that means (dp)²/dx, and that just isn't right. d(p²)/dx, on the other hand, does make sense, and d(p²) can be integrated over using variable substitution.

$$\int f(u)du = \int f(x)\frac{du}{dx}dx$$

Which in this case, turns integral over d(p²) into integral over 2p dp.

12. Aug 26, 2011

### tiny-tim

hi hermano!

(btw, we say "a fault in the mathematics" … though you can say "a crime against mathematics"! )
No, we often treat a derivative exactly like a fraction, and multiply both sides by dx (and then integrate) …

that's fine
∫ d(p2) = [p2]

13. Aug 26, 2011

### hermano

Re: intgral of dp^2??

Thanks, now it is clear for me!

Maybe you both can help me with another small issue I'm facing with this equation! For the integral of my pressure, I have to fill in the upper and lower boundary for p^2! However, I doubt of the input pressure is my upper or lower boundary, and also the same for the output pressure. What tells me which pressure (input - output) is the upper and lower boundary for my integral?

The equation gives the mass flow through a pipe (Poiseuille) from the high input pressure to the lower ouput pressure.