Question about apply integrals in finding volume of a pyramid?

Question

The answers entered already is what I got but either one or both are wrong.

Can someone help me solve this problem?

Answer 1

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.

Answer 2

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)