Question about apply integrals in finding volume of a pyramid? 
The answers entered already is what I got but either one or both are wrong.
Can someone help me solve this problem?
 A: The first entry is right, if you choose $y=0$ at the bottom of the pyramid. So the volume is 
$$\int_0^{10}\frac{1}{4}(10-y)^2\,dy.$$
There must have been slippage in the evaluation of the integral, it should be $\frac{250}{3}$.
Remark: I would prefer to assume that the pyramid is pretty light, and turn it upside down, with the "apex" at the origin. Then the area of cross-section is $\frac{1}{4}y^2$, somewhat more pleasant. 
A: Consider the triangle whose base is a midline of the square base, parallel to an edge, and whose vertex is that of the pyramid.  Take a section at height  y from the base, and let its lenghth be  u.  Then its distance down from the top is  10 - y;  By similar triangles,
u / 5  =  (10 - y) / 10.
Then the width of the square is (10 - y) / 2
Its area (cross-section) is (10 - y)^2 / 4
So the volume of a thin slice, of thickness  δx,  (10 - y)^2 / 4 * δx.
Sum, and take limits as δx → 0:
The volume of the pyramid is
∫ [( (10 - y)^2 / 4 dy  for 0 ≤ y ≤ 10]
= [-(10 - y)^3 / 12 : 0 ≤ y ≤ 10]
= (1/12)*(1000 - 0)
= 250/3  =  83.3333...
(Oops!  mixed up y and x first go!  if you saw it)
