# [SOLVED] volumes of square based pyramid

• Feb 22nd 2009, 05:50 PM
razorfever
[SOLVED] volumes of square based pyramid
find the volume of a pyramid, whose base is a square of side b and height is h

consider a partition of the x-axis and subdivide into this slices with square cross-sections

i can do this with partitions of the y-axis but is this question i have to do it with partitions of the x-axis

i cant seem to do this, can someone help me out and write down the expression for the volume explaining how they got each part
i've been trying this for hours
• Feb 22nd 2009, 06:06 PM
skeeter
sketch the lines ...

$y = \frac{b}{2h} x$ and $y = -\frac{b}{2h} x$

for $x \geq 0$ to see a "side" view of the pyramid.

$V = \int_0^h \left[\frac{b}{2h} x - (-\frac{b}{2h} x)\right]^2 \, dx$

$V = \int_0^h \left(\frac{b}{h}\right)^2 x^2 \, dx$

$V = \left(\frac{b}{h}\right)^2 \left[\frac{x^3}{3}\right]_0^h$

$V = \frac{b^2}{h^2}\left[\frac{h^3}{3} - 0\right] = \frac{b^2 h}{3}$
• Feb 22nd 2009, 07:27 PM
razorfever
in the first step why do you subtract the two values and then square the whole thing?
what i was doing was just squaring the first part without subtracting the second part like you did
can you explain why you subtracted
thanks for the help
i'm not that good with this..
• Feb 23rd 2009, 07:32 AM
skeeter
Quote:

Originally Posted by razorfever
in the first step why do you subtract the two values and then square the whole thing?

the vertical distance between the two lines is the "base" length at any height of the pyramid.

$\frac{b}{2h}x - \left(-\frac{b}{2h}x \right) = \frac{b}{h}x$

at $x = h$, length of the base side = $b$