Integration by Parts: With Partials

[Saladsamurai]

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:

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

Homework Equations

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

The Attempt at a Solution

Let

$$\mu = v(x,y)<br /> \Rightarrow d\mu = <br /> \frac{\partial{v}}{\partial{x}}\,dx +<br /> \frac{\partial{v}}{\partial{y}}\,dy \qquad(3)$$

And let

$$d\theta =<br /> \frac{\partial{u}}{\partial{y}}\,dy \qquad(4)$$

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 =<br /> \frac{\partial{u}}{\partial{y}}\,dy <br /> [/tex]<br /> <br /> it looks as though I have assumed that θ=θ(y) alone. Does this help me at all? Can I now say that <br /> <br /> $$d\theta =<br /> \frac{\partial{u}}{\partial{y}}\,dy <br /> =\frac{d\,u}{d\,y}d\,y=\,du \qquad(5)$$<br /> <br /> ?[/itex]

 

[hunt_mat]

It's just like normal:

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