# Compute Integral of Function z∂u(z)/∂z

• crocomut
In summary: In other words you have to consider the total differential: this is written in the following form for a three dimensional function that maps to a one dimensional output (you can extend the idea to more dimensions if you want):du = ∂u/∂x dx + ∂u/∂y dy + ∂u/∂z dz, where du is the infinitesimal change in u and the others are as expected.From this it should appear the case that unless you assume your derivatives to be zero for the other variables, that you can't just get u from one partial derivative.In fact I would say that the way the author did it was also misleading as well. You said
crocomut
How do I compute the integral

∫z ∂u(z)/∂z dz

Thanks.

crocomut said:
How do I compute the integral

∫z ∂u(z)/∂z dz

Thanks.

Hey crocomut and welcome to the forums.

Are you aware of integration by parts?

chiro said:
Hey crocomut and welcome to the forums.

Are you aware of integration by parts?

Yes, this is what I attempted and I got

zu - ∫udz

Does it make sense?

crocomut said:
Yes, this is what I attempted and I got

zu - ∫udz

Does it make sense?

It does but I'm wondering what kind of context this question is in?

For example are you asked to find it out in terms of the function u? Are you given u or the partial derivative? Are you given the integral of u with respect to z and thus have to find out the integral given only this?

If you have this kind of information it would be helpful for us to know this since this context helps dictate what kind of answer you should provide.

chiro said:
It does but I'm wondering what kind of context this question is in?

For example are you asked to find it out in terms of the function u? Are you given u or the partial derivative? Are you given the integral of u with respect to z and thus have to find out the integral given only this?

If you have this kind of information it would be helpful for us to know this since this context helps dictate what kind of answer you should provide.

This is not a homework question or anything like that, I am just trying to learn about the log-law of the wall in fluid dynamics.

So in reality what I have to do is solve for u in the following expression (there are constants in there but I removed for clarity):

∂u/∂z = 1/z

The way this is solved in the book is u = ln(z) + C, so they integrate the left hand side and right hand side separately. However, if I solve it another way, i.e. if I put z on the left I get:

z∂u/∂z - 1 = 0

Integrating by parts I get zu - ∫udz - z + C = 0 from which I am not able to arrive at u = ln(z) + C. That is what is confusing me.

Thanks.

crocomut said:
This is not a homework question or anything like that, I am just trying to learn about the log-law of the wall in fluid dynamics.

So in reality what I have to do is solve for u in the following expression (there are constants in there but I removed for clarity):

∂u/∂z = 1/z

The way this is solved in the book is u = ln(z) + C, so they integrate the left hand side and right hand side separately. However, if I solve it another way, i.e. if I put z on the left I get:

z∂u/∂z - 1 = 0

Integrating by parts I get zu - ∫udz - z + C = 0 from which I am not able to arrive at u = ln(z) + C. That is what is confusing me.

Thanks.

The thing you have to be careful about is the dimension of your u function.

In your function you have indicated a partial derivative which implies that your function u is not just a function of z, but of possibly other variables. I'm assume since you are studying fluids, that it is at least a function of x and y as well.

In other words you have to consider the total differential: this is written in the following form for a three dimensional function that maps to a one dimensional output (you can extend the idea to more dimensions if you want):

du = ∂u/∂x dx + ∂u/∂y dy + ∂u/∂z dz, where du is the infinitesimal change in u and the others are as expected.

From this it should appear the case that unless you assume your derivatives to be zero for the other variables, that you can't just get u from one partial derivative.

In fact I would say that the way the author did it was also misleading as well. You said you took out the constants, but in this case they are probably very important.

My guess is that the constants represent information about the other parameters like the x and y parameters if they exist. So if your solution is C1log(z) + C2 then C1 and C2 will involve your other variables if they are not just numeric constants.

zu - ∫udz - z = zu - z(log(z) - 1 + C) - z = zu - zlog(z) + z - z - Cz = zu - zlog(z) = z(u - log(z)) = Cz so your extra z term cancels out and gives you u - log(z) = C which gives u = log(z) + C: Remember that you substitute u into the integral including the C term and you don't include an extra C term before you integrate. Keep in mind that constant of integration for u is actually the C you are given in u = log(z) + C: not being aware of this will confuse you into why I don't add an extra C term when I did my calculation above.

chiro said:
The thing you have to be careful about is the dimension of your u function.

In your function you have indicated a partial derivative which implies that your function u is not just a function of z, but of possibly other variables. I'm assume since you are studying fluids, that it is at least a function of x and y as well.

In other words you have to consider the total differential: this is written in the following form for a three dimensional function that maps to a one dimensional output (you can extend the idea to more dimensions if you want):

du = ∂u/∂x dx + ∂u/∂y dy + ∂u/∂z dz, where du is the infinitesimal change in u and the others are as expected.

From this it should appear the case that unless you assume your derivatives to be zero for the other variables, that you can't just get u from one partial derivative.

In fact I would say that the way the author did it was also misleading as well. You said you took out the constants, but in this case they are probably very important.

My guess is that the constants represent information about the other parameters like the x and y parameters if they exist. So if your solution is C1log(z) + C2 then C1 and C2 will involve your other variables if they are not just numeric constants.

zu - ∫udz - z = zu - z(log(z) - 1 + C) - z = zu - zlog(z) + z - z - Cz = zu - zlog(z) = z(u - log(z)) = Cz so your extra z term cancels out and gives you u - log(z) = C which gives u = log(z) + C: Remember that you substitute u into the integral including the C term and you don't include an extra C term before you integrate. Keep in mind that constant of integration for u is actually the C you are given in u = log(z) + C: not being aware of this will confuse you into why I don't add an extra C term when I did my calculation above.

Thank you for your time, I think I was correct in the way I was thinking and just needed confirmation. Your answer was great.

## 1. What is the definition of a compute integral?

A compute integral is a mathematical process used to find the area under a curve in a given interval. It involves breaking down the area into smaller, known shapes and summing them together to find the total area.

## 2. What is the difference between a definite and indefinite integral?

A definite integral has specific limits of integration, meaning it calculates the area under a curve within a specific interval. An indefinite integral does not have limits of integration and is used to find the general formula for an integral.

## 3. How is the integral of a function z∂u(z)/∂z computed?

The integral of a function z∂u(z)/∂z is computed by first finding the anti-derivative of the function. This is done by reversing the process of taking the derivative, and the resulting function is then evaluated at the given limits of integration.

## 4. What is the purpose of computing the integral of a function?

The purpose of computing the integral of a function is to find the area under a curve, which can have many real-world applications. It can also be used to find the volume of a three-dimensional shape or to solve problems in physics and engineering.

## 5. Are there any techniques or shortcuts for computing integrals?

Yes, there are several techniques and shortcuts for computing integrals, such as substitution, integration by parts, and partial fraction decomposition. These techniques can simplify the process of finding the anti-derivative and make computing integrals more efficient.

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