# Integration help

I am having difficulty in integrating this one. My objective is to find height as a function of time; however each time I actually do the integration, I end up with being able to find height as a function of both time and height. I have attempted using Maple to find a unique solution but it produces a (Lambar delta) (possible spelling error). Here is my integration.

int(t)=int([8(mu)(h)]/[2(sigma)(g)(r)-(rho)(g)(h)(r^2)] dh

I keep getting non real answers when I plug in values. Sigma is a surface tension but it keeps getting caught into natural logs with a negative value(impossible)

Any help would be quite appreciated.

arildno
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What are you trying to do here??
Is it to find the anti-derivative:
$$I=\int\frac{8\mu{h}}{2\sigma{gr}-\rho{gh}r^{2}}dh$$
If so, what are constants, and what is variable quantities with respect to which variables?

the only variables are time and height. The others are constants.

arildno
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What is r?

I will define all variables
t=time
h=height
mu= fluid viscocity
sigma=surface tension
g=gravitational force
rho= density

arildno
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Well, and how does t enter the picture.

Whatever do you mean by int(t)??

the integral of t from t=0 to t so basically the left hand side of the equation reduces to t.

arildno
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The expression $\int_{0}^{t}tdt$????
First of all, your notation is a meaningless conjunction of integration variable and limit of integration, nor would it in this abuse of notation equal t.

Are you sure your left-hand side shouldn't be $\int_{0}^{t}d\tau=t$?

right I forgot to say with what respect I was integrating the left hand side, but I am sure that the left hand side reduces to t.

I agree with the first equation that you posted except the left hand side should be t instead of I. the limits of integration from the right side are h= 0 to h=h(final)

arildno
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Okay, so what your ACTUAL equation should be is:
$$t=\int_{0}^{h(t)}\frac{8\mu{H}}{2\sigma{gr}-\rho{gH}r^{2}}dH$$

That, at the very least, is a mathematically meaningful expression.
Is that what you meant?

yes. I'm sorry, but I don't know how to use the equation editor

arildno
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Now, you may rewrite your right-hand side as:
$$t=K\int_{0}^{h(t)}\frac{h}{1-CH}dH$$
for suitable constants K, C.
Can you perform that integration?

I'm sorry I don't guess I follow you. That isn't something that I have seen in my undergraduate mathematics.

arildno
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Sorry, it should have been:

$$t=K\int_{0}^{h(t)}\frac{H}{1-CH}dH$$
for suitable constants K, C.

That's just algebra.

okay, now I see. Thanks. I can do that integration quite easily.

arildno