Question

Quick question! @eSpeX @amistre64

When I calculate a rectangular prism's volume that is measured in inches or centemeters with three given measurements after I calculate in the answer do I put cm^3 or ??

Answer 1

Anyone know?? l x w x h = cm^3 or ?? if the prism is measured in inches or squared? I know dumb question but need advice.

Answer 2

if your measurements are in inches, your result will be in in^3.

say l = 2 in, w = 3 in, h = 4 in
V = l * w * h = (2 in)(3 in)(4 in) = 2*3*4 * in*in*in = 24 in^3

Answer 3

because there are three given numbers?

Answer 4

keep the units with the numbers, and treat them as if they have a value of 1 (so they don't change the answer).

Answer 5

Sooooo thankful!!!!!!!

Answer 6

is this the same for all measurements?

Answer 7

it's also very helpful when doing conversions to do them as fractions. Say you had one of those measurements in feet instead of inches:

l = 2 ft, w = 3 in, h = 3 in
\[V = 2 ft * 3 in * 3in = 2 \cancel{ft} * (12 in/1 \cancel{ft}) * 3 in * 3 in = 216 in^3 \]

Answer 8

If you accidentally set up the conversion backwards, like this:

\[V = 2 ft * 3 in * 3in = 2 ft * (1 ft/12 in) * 3 in * 3 in = 1.5 ft^2 in^2 \]and you're scratching your head saying "what kind of a unit of volume is that?" which is your clue that you did it wrong :-)

Answer 9

ok so if you ahve multiple measurements cancel out for everyone until you have only one type and use it. Did I make sense?

Answer 10

yes, but make sure that they really cancel out, not just cancel out because you want them to cancel out :-)

Answer 11

all the ones I have are the same though.

Answer 12

that's fine. just do the algebra honestly, make sure the units make sense, and that means your answer was probably done correctly (assuming you didn't make a mistake in the arithmetic).

Answer 13

ok. You make sense. Thanks. : ) I guess it just seemed weird to me to put cm^3 or in^3 to everyone and including the surface area problems.

Answer 14

well, surface area problems are going to be in^2 or cm^2 or whatever. The underlying idea is called dimensional analysis. There are actually some rather complicated things you can figure out just by reasoning on what the units must be...

Answer 15

haha I was just changing the surafce area to cm^2 I thought I did that one wrong. caught myself.

Answer 16

okay, good luck!

Answer 17

ok last question. What about for cylinders?

Answer 18

Again, what are the dimensions?

Answer 19

oh wait one more i am having trouble ffiguring out the surface area and volume for a half dollar coin??

Answer 20

coins are mm

Answer 21

put in the units as you compute the result, and you'll get the right units.

\[V_{cyl} = 4\pi r^2 h\]\[V = 4 \pi i L^2 * L = 4\pi L^3\] L being (length) so your units will be of the form (length)^3

Answer 22

I just have them in mm for coins

Answer 23

okay, mm is a unit of length, right? it's not temperature, or mass, or anything else, it's length. no matter what the numbers are, volume will always have dimensions of (length)^3 . Area will always be (length)^2.

Answer 24

Like I said, instead of putting (2) in your formula if the problem says the dimension is 2 mm, you put (2 mm) in the formula and treat the "mm" just as you would an algebraic variable. What you get out at the end will be the proper unit.

Answer 25

volume represents a 3d construct in R^3

Answer 26

so as long as your not trying to mix unit measures of inches and centimeters and kilometers ... cubing is fine

Answer 27

Area represents a R^2 map

Answer 28

ok Make sense. Thanks @whpalmer4 and thanks for the extra info too @amistre64

I writing all the notes down.