Why is Earth not a sphere?
Normally in the absence of rotation, the natural tenancy of gravity is to pull the Earth together in the shape of a sphere.

However the Earth in fact bulges at the equator, and the diameter across the equatorial plane is 42.72 km more than the diameter from pole to pole.

This is due to the rotation of the Earth.  



As we can see in the image above, the spinning disk appears to bulge at the points on the disk furthest from the axis of rotation. 

This is because in order for the particles of the disk to remain in orbit, there must be an inward force, known as the centripetal force, given by:

$$F = \frac{mv^2}{r},$$

where $F$ is the force, $m$ is the mass of rotating body, $v$ is the velocity and $r$ is the radius of particle from the axis of rotation.

If the disk is rotating at a given angular velocity, say $\omega$, then the tangential velocity $v$, is given by $v = \omega r$.

Therefore,

$$F = m\omega^2r$$

Therefore the greater the radius of the particle, the more force is required to maintain such an orbit.

Therefore particles on the Earth near the equator, which are farthest from the axis of rotation, will buldge outwards because they require a greater inward force to maintain their orbit.



Additional details for more mathematically literate now that mathjax is enabled:

The net force on an object rotating around the equator with a radius $r$ around a planet with a gravitational force of $\frac{Gm_1m_2}{r^2}$ is the centripetal force given by,

$$F_{net} = \frac{Gm_1m_2}{r^2} - N = m\omega^2r,$$ where $N$ is the normal force.

Re-arranging the above equation gives:

$$N = \frac{Gm_1m_2}{r^2} - m\omega^2r$$

The normal force here is the perceived downward force that a rotating body observers.  The equation shows that the perceived downward force is lessened due to the centripetal motion.  The typical example to illustrate this is there is an appearance of 0 gravity in a satellite orbiting the Earth, because in this situation the centripetal force is exactly balanced by the gravitational force.  On Earth however, the centripetal force is much less than the gravitational force, so we perceive almost the whole contribution of $mg$.

Now we will examine how the perceived gravitational force differs at different angles of latitude.  Let $\theta$ represent the angle of latitude.  Let $F_G$ be the force of gravity.

In vector notation we will take the $j$-direction to be parallel with the axis of rotation and the $i$-direction to be perpendicular with the axis of rotation.

In the absence of the Earth's rotation,

$$F_G = N = (-\frac{Gm_1m_2}{r^2}\cos\theta)\tilde{i} + (-\frac{Gm_1m_2}{r^2}\sin\theta)\tilde{j}$$

It is easily seen that the above equation represents the perceived force of gravity in the absence of rotation.  Now the centripetal force acts only in the i-direction, since it acts perpendicular to the axis of rotation.

If we let $R_{rot}$ be the radius of rotation, then the centripetal force is $m_1\omega^2R_{rot}$, which for an angle of latitude of $\theta$ corresponds to $m_1\omega^2r\cos{\theta}$

$$N = (-\frac{Gm_1m_2}{r^2} + m_1\omega^2r)\cos{\theta}\tilde{i} + (-\frac{Gm_1m_2}{r^2})\sin{\theta}\tilde{j}$$

By comparing this equation to the case shown earlier in the absence of rotation, it is apparent that as $\theta$ is increased (angle of latitude), the effect of rotation on perceived gravity becomes negligible, since the only difference lies in the $x$-component and $\cos\theta$ approaches 0 as $\theta$ approaches 90 degrees latitude.  However it can also be seen that as theta approaches 0, near the equator, the $x$-component of gravity is reduced as a result of the Earth's rotation.  Therefore, we can see that the magnitude of $N$ is slightly less at the equator than at the poles.  The reduced apparent gravitational pull here is what gives rise to the slight bulging of the Earth at the equator, given that the Earth was not originally as rigid as it is today (see other answer).