Partial integration of the function f = 1

  • Thread starter caleb5040
  • Start date
  • #1
22
0
Given a function z(x,y), is the following expression valid:

∫(1)∂z = z

Does this even make sense? I came up with this in my differential equations class, but I'm not sure if it actually means anything.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Correct except that you do not have the "constant of integration". Which, because the partial derivative with respect to z treats other variables as constants, may be a function of those other variables. That is, if your variables are x, y, and z, the "anti-derivative" of 1, with respect to z, is [itex]z+ f(x,y)[/itex] where f can be any function of two variables.
 
  • #3
22
0
I think that makes sense. Thanks!
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Note that if you have the three partial differential equations
[tex]\frac{\partial f}{\partial x}= 1[/tex]
[tex]\frac{\partial f}{\partial y}= 1[/tex]
[tex]\frac{\partial f}{\partial z}= 1[/tex]

The last equation gives, as I said, f(x,y,z)= z+ g(x,y). Differentiating that with respect to y,
[tex]\frac{\partial f}{\partial y}= \frac{\partial g}{\partial y}= 1[/tex]
so that g(x,y)= y+ h(x). Differentiating that with respect to x,
[tex]\frac{\partial g}{\partial x}= \frac{dh}{dx}= 1[/tex]
which gives h(x)= x+ C. Putting all of those together, f(x,y,z)= z+ g(x,y)= z+ y+ h(x)a= z+ y+ x+ C.
 

Related Threads on Partial integration of the function f = 1

Replies
2
Views
1K
Replies
5
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
10
Views
2K
Top