Integration by Parts: With Partials

  • #1
Saladsamurai
3,019
6

Homework Statement



I don't know why, but the partials are really confusing me here. I need to integrate the following expression in a derivation:

[tex]I = \int_0^\delta v(x,y)\frac{\partial{u(x,y)}}{\partial{y}}\,dy \qquad(1)[/tex]


Homework Equations



I am supposed to integrate by parts here. [itex]\int \mu\,d\theta = \mu\theta - \int\theta\,d\mu \qquad(2)[/itex]

The Attempt at a Solution



Let

[tex]\mu = v(x,y)
\Rightarrow d\mu =
\frac{\partial{v}}{\partial{x}}\,dx +
\frac{\partial{v}}{\partial{y}}\,dy \qquad(3)[/tex]

And let

[tex]d\theta =
\frac{\partial{u}}{\partial{y}}\,dy \qquad(4)[/tex]

Now I am really not sure what to do with these quantities. So let me state some questions here:

I) Is this the best choice for my µ and dθ?

II) Since I have assumed that

[itex]d\theta =
\frac{\partial{u}}{\partial{y}}\,dy
[/tex]

it looks as though I have assumed that θ=θ(y) alone. Does this help me at all? Can I now say that

[tex] d\theta =
\frac{\partial{u}}{\partial{y}}\,dy
=\frac{d\,u}{d\,y}d\,y=\,du \qquad(5)[/tex]

?
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,760
27
It's just like normal:

[tex]
\int_{0}^{\delta}v\frac{\partial u}{\partial y}dy=\left[ uv\right]_{0}^{\delta}-\int_{0}^{\delta}u\frac{\partial v}{\partial y}dy
[/tex]
 

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