It's easy to find the volume of a cone, but how would you find the volume of a pyramid using integration? With height h and base length a. And if it's a SQUARE BASED pyramid.

Also...what's the antiderivative of (e^x)^2?

Printable View

- Jul 9th 2008, 08:53 PMSportfreundeKeaneKentFinding Volume of Pyramid (Integration)?
It's easy to find the volume of a cone, but how would you find the volume of a pyramid using integration? With height h and base length a. And if it's a SQUARE BASED pyramid.

Also...what's the antiderivative of (e^x)^2? - Jul 9th 2008, 09:40 PMmr fantastic
1. Read this: Calculus/Volume - Wikibooks, collection of open-content textbooks

2. Read this thread: http://www.mathhelpforum.com/math-he...functions.html - Jul 9th 2008, 10:41 PMReckoner
- Jul 9th 2008, 11:10 PMticbol
Here is one way.

We divide the pyramid into 4 congruent right triangular pyramids.

Meaning, cut the square-based pyramid along the two diagonals of the square base, vertically. (If you understand that, you're good (Happy)).

If the square base is $\displaystyle a$ by $\displaystyle a$, then each diagonal is sqrt(2)*$\displaystyle a$ long.

Since the diagonals of a square are perpendicular and they bisect each other, then anyone of those 4 congruent right triangular pyramids has a base that is a right triangle whose hypotenuse is $\displaystyle a$ and whose legs are both sqrt(2)$\displaystyle a$/2 = $\displaystyle a$/sqrt(2) each.

We get the volume of one of these right triangular pyramids, multiply that by 4, and we get the volume of the right square-based pyramid.

In one of the right triangular pyramids:

Set the dV ---the element of the volume---anywhere from y=0 and y=h.

>>>dA = area of the dV

Say dV is a y below the apex, and the hypotenuse of the dA is x.

So, the area of the dA is (1/2)(x /srt(2))(x /sqrt(2)) = (x^2)/4

And the dV is [(x^2)/4]dy

-------------------------------------------------

We find the relationship of y and x.

In one vertical face of the right triangular pyramid, there are two right triangles;

Bigger right triangle has base = $\displaystyle a$/sqrt(2), and height = h.

Smaller right triangle has base = x/sqrt(2), and height = y.

By proportion,

[$\displaystyle a$/sqrt(2)] /h = [x/sqrt(2) /y]

Cross multiply,

$\displaystyle a$*y = h*x

So, x = ($\displaystyle a$/h)y

------------------

Hence, dV = [(x^2)/4]dy = (1/4)[$\displaystyle a$/h)^2 *y^2]dy

dV = (1/4)($\displaystyle a$^2 /h^2)[y^2]dy

And so, for the whole right square-based pyramid,

V = 4{(1/4)($\displaystyle a$^2 /h^2)INT(0 to h)[y^2]dy} ...going down is positive.

V = ($\displaystyle a$^2 h^2)INT(0 to h)[y^2]dy

V = ($\displaystyle a$^2 /h^2)[(y^3)/3](0 to h)

V = ($\displaystyle a$^2 /h^2)[(h^3)/3 -0]

V = (1/3)($\displaystyle a$^2)h

That is it. - Jul 10th 2008, 07:36 AMSportfreundeKeaneKent
I don't really understand the first part ticbol...is there a diagram to explain it?

What I'm also trying to do is explain why the cross sections parallel to the base will be A(x)= (a^2/h^2)*x^1 - Jul 10th 2008, 07:50 AMgalactus
Yes, here is a diagram I made best I could in Paint. Hope it helps.

The cross sections perpendicular to the y-axis(parallel to the x-axis) are squares. If s is a side length of this square then:

$\displaystyle \frac{\frac{s}{2}}{\frac{a}{2}}=\frac{h-y}{h}$

or

$\displaystyle s=\frac{a}{h}(h-y)$

$\displaystyle \text{area of square}=A(y)=s^{2}=\frac{a^{2}}{h^{2}}(h-y)^{2}$

$\displaystyle V=\int_{0}^{h}A(y)dy=\int_{0}^{h}\frac{a^{2}}{h^{2 }}(h-y)^{2}dy=\frac{1}{3}a^{2}h$