Integration of partial derivative

In summary, the integration of partial derivatives is the process of finding a function from its derivatives, and it is used to determine the original function when only the derivatives are known. It differs from integration of single-variable functions as it involves multiple independent variables and requires the use of multivariable calculus techniques. This has various applications in science, specifically in physics and engineering, and can be used to solve real-world problems such as determining velocity and acceleration, finding optimal solutions, and calculating electric or magnetic fields. Some commonly used techniques for integration of partial derivatives include separation of variables, substitution, and integration by parts.
  • #1
acg8934
4
0

Homework Statement


Given a body in a state of plane stress with no body forces where
[tex]\sigma[/tex]x=x2y
[tex]\sigma[/tex]y=(y3-3y)/3

Find [tex]\tau[/tex]xy


Homework Equations


For plane stress
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y + X = 0

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x + Y = 0

No body forces: X=0 Y=0

Simplified:
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x

The Attempt at a Solution



[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = 2xy

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = y2-1

Putting this into above equations gives:
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y = -2xy
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x = -y2+1

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-2xy[tex]\partial[/tex]y
[tex]\tau[/tex]xy = -y2x

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-y2+1[tex]\partial[/tex]x
[tex]\tau[/tex]xy = -y2x + x

These should both be equal, but I'm not sure what I'm doing wrong. I think its my integration.
Please help.
 
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  • #2
acg8934 said:

Homework Statement


Given a body in a state of plane stress with no body forces where
[tex]\sigma[/tex]x=x2y
[tex]\sigma[/tex]y=(y3-3y)/3

Find [tex]\tau[/tex]xy


Homework Equations


For plane stress
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y + X = 0

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y + [tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x + Y = 0

No body forces: X=0 Y=0

Simplified:
[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = -[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x

The Attempt at a Solution



[tex]\partial[/tex][tex]\sigma[/tex]x/[tex]\partial[/tex]x = 2xy

[tex]\partial[/tex][tex]\sigma[/tex]y/[tex]\partial[/tex]y = y2-1

Putting this into above equations gives:
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]y = -2xy
[tex]\partial[/tex][tex]\tau[/tex]xy/[tex]\partial[/tex]x = -y2+1

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-2xy[tex]\partial[/tex]y
[tex]\tau[/tex]xy = -y2x

[tex]\int[/tex][tex]\partial[/tex][tex]\tau[/tex]xy = [tex]\int[/tex]-y2+1[tex]\partial[/tex]x
[tex]\tau[/tex]xy = -y2x + x

These should both be equal, but I'm not sure what I'm doing wrong. I think its my integration.
Please help.

You forgot that when you do partial integration, say with respect to y, you have to add an arbitrary function of x. Similarly when you integrate partially with respect to x you have to add an arbitrary function of y. This will give you the necessary freedom to determine your unknown function.
 
  • #3
Ok, but doing that gives me

[tex]\tau[/tex]xy=-y2x + f(x) for the first

and

[tex]\tau[/tex]xy=-y2x + x + f(y)

Is that correct? If so how do I figure out what the actual [tex]\tau[/tex]xy is?

Thanks
 
  • #4
Don't use f as both a function of x alone and another function of y alone.

So you have T(x, y) = -y^2x + f(x), and
T(x, y) = -y^2x + x + g(y).
Edit: added x in the line above.

Does that help?

My LaTeX doesn't seem to be rendering correctly, so I switched tau to T.
 
Last edited:
  • #5
How do I find f(x) and g(y)?

I know the tau's should be equal.
 
  • #6
I omitted x from the 2nd representation by accident. Both forms of T(x, y) have -y2x. The first representation doesn't have any functions of y alone, so g(y) = ? The 2nd representation has x, so f(x) = ?
 
  • #7
So does that mean f(x)=x and g(y)=0? So Txy=-y^2x+x?
 
  • #8
Almost. g(y) = C, a constant. So T(x, y) = -xy^2 + x + C.
 

1. What is the definition of integration of partial derivatives?

The integration of partial derivatives is a mathematical process that involves finding a function from its derivatives. It is the inverse operation of differentiation, and it is used to determine the original function when only the derivatives are known.

2. How is integration of partial derivatives different from integration of single-variable functions?

Integration of partial derivatives is different from integration of single-variable functions because it involves finding a function from its partial derivatives with respect to multiple independent variables. This requires the use of multivariable calculus techniques, such as the chain rule and partial integration.

3. What are the applications of integration of partial derivatives in science?

The integration of partial derivatives has many applications in science, particularly in physics and engineering. It is used to solve problems involving rates of change, optimization, and motion of objects in multiple dimensions. It is also used in the study of fluid mechanics, electromagnetism, and thermodynamics.

4. Can integration of partial derivatives be used to solve real-world problems?

Yes, integration of partial derivatives is commonly used to solve real-world problems in various fields. For example, it can be used to determine the velocity and acceleration of a moving object, to find the optimal solution to a production or manufacturing problem, or to calculate the electric or magnetic field at a specific point in space.

5. What are some common techniques for integration of partial derivatives?

Some common techniques for integration of partial derivatives include the method of separation of variables, the method of substitution, and the method of integration by parts. These techniques are used to simplify the integration process and make it easier to solve complex problems involving multiple variables.

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