Integration of partial derivative

[acg8934]

Homework Statement

Given a body in a state of plane stress with no body forces where
\sigmax=x²y
\sigmay=(y³-3y)/3

Find \tauxy

Homework Equations

For plane stress
\partial\sigmax/\partialx + \partial\tauxy/\partialy + X = 0

\partial\sigmay/\partialy + \partial\tauxy/\partialx + Y = 0

No body forces: X=0 Y=0

Simplified:
\partial\sigmax/\partialx = -\partial\tauxy/\partialy

\partial\sigmay/\partialy = -\partial\tauxy/\partialx

The Attempt at a Solution

\partial\sigmax/\partialx = 2xy

\partial\sigmay/\partialy = y²-1

Putting this into above equations gives:
\partial\tauxy/\partialy = -2xy
\partial\tauxy/\partialx = -y²+1

\int\partial\tauxy = \int-2xy\partialy
\tauxy = -y²x

\int\partial\tauxy = \int-y²+1\partialx
\tauxy = -y²x + x

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

 

[AEM]

Homework Statement

Given a body in a state of plane stress with no body forces where
\sigmax=x²y
\sigmay=(y³-3y)/3

Find \tauxy

Homework Equations

For plane stress
\partial\sigmax/\partialx + \partial\tauxy/\partialy + X = 0

\partial\sigmay/\partialy + \partial\tauxy/\partialx + Y = 0

No body forces: X=0 Y=0

Simplified:
\partial\sigmax/\partialx = -\partial\tauxy/\partialy

\partial\sigmay/\partialy = -\partial\tauxy/\partialx

The Attempt at a Solution

\partial\sigmax/\partialx = 2xy

\partial\sigmay/\partialy = y²-1

Putting this into above equations gives:
\partial\tauxy/\partialy = -2xy
\partial\tauxy/\partialx = -y²+1

\int\partial\tauxy = \int-2xy\partialy
\tauxy = -y²x

\int\partial\tauxy = \int-y²+1\partialx
\tauxy = -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.

 

[acg8934]

Ok, but doing that gives me

\tauxy=-y²x + f(x) for the first

and

\tauxy=-y²x + x + f(y)

Is that correct? If so how do I figure out what the actual \tauxy is?

Thanks

 

[Mark44]

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.

 

[acg8934]

How do I find f(x) and g(y)?

I know the tau's should be equal.

 

[Mark44]

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) = ?

 

[acg8934]

So does that mean f(x)=x and g(y)=0? So Txy=-y^2x+x?

 

[Mark44]

Almost. g(y) = C, a constant. So T(x, y) = -xy^2 + x + C.