Current density? The current $i$ can be defined as:
$$i = \int \vec{J} \dot{}d\vec{A} $$
where $\vec{J}$ is the current density and $d\vec{A}$ is the area vector.
Is it possible for:
$$i = \int \vec{J} \dot{}d\vec{A} \neq J\times A $$
where $A$ is the area? 
My book gives the example of when the current is not parallel to the area vector. However, I don't see why the direction of the electron flow matters. If 3 electrons passes through a circle per second, wouldn't the current be 3e/s regardless of whether they pass at 45 degrees or 90, since they pass regardless?
 A: (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 it can be that
$$i = \int \vec{J} \dot{}d\vec{A} \neq J\times A $$
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.
Edit: Of course, if the area is perpendicular to the current density then I=j*A.
A: 
If 3 electrons passes through a circle per second, wouldn't the
  current be 3e/s regardless of whether they pass at 45 degrees or 90,
  since they pass regardless?

It would. But here we know that there is a stream of electrons going at a certain speed $v$ having a cross sectional area $A$. Then we also know that our surface has an area $a$. ($a<A$). But, it is not necessary that our surface will intercept all the electrons within the area $a$ making the current $av$. This is because the number of electrons it intercepts also depends on how our area is oriented with respect to the electron beam. If the plane of our surface is the parallel to the surface of the electron beam, then it will indeed have a current of $av$ as $av$ electrons cross the surface in unit time. But if it say oriented at some angle $\theta$ with the direction of the beam. Then the part of beam that passes through the area will now no longer be $a$, but it will be $a\cos\theta$ because that is the area of the beam that is intercepted by our differently oriented surface. Thus, our surface of area $a$ intercepts and area $a\cos\theta$ of the beam and hence the number of electron crossing the surface will be $a\cos\theta v$. Here also the number of electrons crossing per unit time is the current irrespective of how they cross the surface, but due to different orientation of our area, some smaller portion of the beam is intercepted and this portion depends on the geometry of our surface's orientation and the direction of the beam. The simple multiplication assumes that the entire area of our surface intercepts an equal area of the electron beam, but this is not the case always.  
The surface integral helps find out the effective area of the beam intercepted which will give the number of electrons crossing, and irrespective of the direction of these electrons, they will constitute the current through the surface.
A: Current density gives an idea about current through a cross section per unit area. If no area is given for current to pass through, then current will be zero. As like no water flows through the ring when held parallel to its flow.  

However, I don't see why the direction of the electron flow matters.  

Remeber current density direction depends on the direction of area vector and not the direction of electrons (actually electron current is not vector quantity). Its the direction of area vector which matters. Garyp has done well in explaining your problem.
