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?
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
2. Read this thread: http://www.mathhelpforum.com/math-he...functions.html
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 $\displaystyle a$ 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
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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
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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.
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
$\displaystyle s=\frac{a}{h}(h-y)$
$\displaystyle \text{area of square}=A(y)=s^{2}=\frac{a^{2}}{h^{2}}(h-y)^{2}$
$\displaystyle V=\int_{0}^{h}A(y)dy=\int_{0}^{h}\frac{a^{2}}{h^{2 }}(h-y)^{2}dy=\frac{1}{3}a^{2}h$