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?

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- 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 by , then each diagonal is sqrt(2)* 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 and whose legs are both sqrt(2) /2 = /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 = /sqrt(2), and height = h.

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

By proportion,

[ /sqrt(2)] /h = [x/sqrt(2) /y]

Cross multiply,

*y = h*x

So, x = ( /h)y

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

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

dV = (1/4)( ^2 /h^2)[y^2]dy

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

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

V = ( ^2 h^2)INT(0 to h)[y^2]dy

V = ( ^2 /h^2)[(y^3)/3](0 to h)

V = ( ^2 /h^2)[(h^3)/3 -0]

V = (1/3)( ^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:

or