Question

Linear Approximations and Differentials
a) Use differentials to find a formula for the approximate volume of a thin cylindrical shell with height h, inner radius r, and thickness Δr.
b) What is the error involved in using the formula from part (a)?

Answer 1

considering cylindrical coordinates and just looking at the r,theta plane
since height/z integral is as simple as multiplying by a constant

Answer 2

This is Single Variable Calculus.

Answer 3

let us assume that dr is a very very small number

Answer 4

er, i realized, i know how to find it through integration and not differential...

Answer 5

lol. It's ok, because you tried :-D

Answer 6

should i continue?

Answer 7

I have to do it with differentials. sorry :-(

Answer 8

but thank you for the alternate approach, I will keep it in mind for when I start Calc II :-)

Answer 9

wait, differential is just the delta r thing right?

Answer 10

eh... it's something like f prime x I think?\[f'(x)\]

Answer 11

I'm not sure how to define it 100% but I know generally how to apply differentials I guess

Answer 12

cor convenience sake\[\Delta r = dr\]
ok non calculus approach

you obviously know that the the area of a circle is \[A=\pi R^2\]
and so going back to the picture, the big circle has a radius of r
and the little circle has a radius of r-dr

Answer 13

since you want to find the small area inbetween the two circles
you just subtract areas

now you make the assumption that dr is infinitesmally small
so we also make the assumption that dr^2 ~0

Answer 14

then multiply by height to get volume of the cylindrical shell

Answer 15

Er, Δr is equal to dr?

Answer 16

well, not easy to type a triangle inline

Answer 17

oh, I copypaste the delta symbol from the nets. Δ
but you can also use \Delta which shows up as \(\Delta\)

Answer 18

well im pretty inexperienced with the wonders that is latex

Answer 19

lol I abuse Google in terms of LaTeX :-P

Answer 20

to set up the integral use fact that volume is sum of areas (circumference*height)
\[V = \int\limits_a^b (2 \pi r) (h) dr\]
\[V = \int\limits_R^{R+dr} (2\pi r)(h) dr\]

Answer 21

oh yea, you could put the differential on the outside
the result should still be teh same

Answer 22

@dumbcow I'm in Single Variable Calculus...

Answer 23

...I don't know anything about integrals.
Zilch. :-(