# Math Help - Finding Volume of Pyramid (Integration)?

1. ## Finding 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?

2. Originally Posted by SportfreundeKeaneKent
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?
1. Read this: Calculus/Volume - Wikibooks, collection of open-content textbooks

3. Mr. F, I think you misread

Originally Posted by SportfreundeKeaneKent
Also...what's the antiderivative of (e^x)^2?
$\int(e^x)^2\,dx = \int e^{2x}\,dx$

$=\frac12\int2e^{2x}\,dx=\frac12e^{2x} + C$

4. Originally Posted by SportfreundeKeaneKent
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?
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 ).

If the square base is $a$ by $a$, then each diagonal is sqrt(2)* $a$ 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 $a$ and whose legs are both sqrt(2) $a$/2 = $a$/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 = $a$/sqrt(2), and height = h.
Smaller right triangle has base = x/sqrt(2), and height = y.
By proportion,
[ $a$/sqrt(2)] /h = [x/sqrt(2) /y]
Cross multiply,
$a$*y = h*x
So, x = ( $a$/h)y

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

And so, for the whole right square-based pyramid,
V = 4{(1/4)( $a$^2 /h^2)INT(0 to h)[y^2]dy} ...going down is positive.

V = ( $a$^2 h^2)INT(0 to h)[y^2]dy
V = ( $a$^2 /h^2)[(y^3)/3](0 to h)
V = ( $a$^2 /h^2)[(h^3)/3 -0]
V = (1/3)( $a$^2)h

That is it.

5. 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

6. 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:

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

or

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

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

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