(The others did already a good job I think but I will still give it a shot. ^^)

"@dmckee But by definition the "rain" density is the "amount" of rain that passes through it per unit area. So regardless of the angle of the frame, the net rain that passes through it would be rain density * area. The fact that holding the frame at an angle results in less rain passing through is reflected in a lower rain density (since less rain passes through per unit area)"

Let's stay in that analogy:

The rain density (lets define it as 100 raindrops per area) stays the same, but the amount that changes is the area which is "seen" by the rain.

If you imagine that you are the rain and you see the frame, hold horizontally, from above, you will see the entire frame (which has let's say 1m²). So the amount of you passing through the frame is $\frac{100\ raindrops}{unit~ area}*1m^2=100~raindrops$

If however the frame is slightly tilted, the amount of frame you see is less, let's say 0.5m². Likewise, if you stretch your arm and look at the back of your hand, the amount of hand you see will change when you tilt it.

So for the tilted frame you have $\frac{100\ raindrops}{unit~ area}*0.5m^2=50~raindrops$


Back to electricity:

This is why 

$$i = \int \vec{J} \dot{}d\vec{A} \neq J\times A $$

for the electric current: the electron density stays the same but the area seen be the electron density changes. This is expressed by the dot product which gives a cosine-relation for the orientation of the current to the area.