Calc: Find a definite integral for the mass

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Homework Help Overview

The problem involves calculating the mass of sugar that settles at the bottom of a cylindrical mug filled with tea. The density of the sugar varies with height, given by the formula d(h) = 0.01(8-h). Participants are tasked with writing a definite integral to represent the mass of sugar in the mug.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to consider the volume of individual slices of the cylinder rather than the entire volume. There is confusion regarding the correct formulation of the integral, particularly in how to incorporate the height of the slices and the density function.

Discussion Status

Several participants are actively engaging with the original poster's attempts, providing feedback and clarifying concepts related to the integration of density and volume. There is an ongoing exploration of how to correctly set up the integral, with no clear consensus yet on the final formulation.

Contextual Notes

Participants are navigating through the nuances of integrating a variable density function over a cylindrical volume, questioning the assumptions about the relationship between height and volume in the context of the problem.

calculusisfun
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Homework Statement


A cylinder mug with a 4 centimeter radius and 8 centimeter height is filled with tea. When sugar is added to the tea, the sugar usually settles to the bottom of the cylinder mug. The density of sugar in the tea at a height h centimeter from the bottom of the cylinder mug is given by the following formula: d(h) = 0.01(8-h). [Note: mass = (volume)(density)]

Write a definite integral that gives the mass of sugar in the mug exactly.

2. The attempt at a solution

Volume of the mug: [PLAIN]http://img839.imageshack.us/img839/2599/calc3.png

Mass = volume x density

Mass = [PLAIN]http://img215.imageshack.us/img215/3589/calc4.png

Not sure if I did this right or completely wrong.

Help would be much appreciated. :)
 
Last edited by a moderator:
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hi calculusisfun !:smile:

no, the cup is divided into horizontal slices of height dh,

so inside your volume-times-density integral, you must put the volume of the slice

it'll be a mutliple of dh

(that's where your "dh" comes from! :wink:)

(you've used the volume of the whole cup, which doesn't give the same result)
 
Thanks for responding tiny-tim! :)

So, would it be like this? [PLAIN]http://img80.imageshack.us/img80/9790/calc5.png

I multiplied d(h) by 16 pi h, because 16 pi h represents the height of an individual slice of the cylinder.

Am I still doing this completely wrong? :p

I feel like such a nerd staying in on Halloween haha, but any help is much appreciated. :)
 
Last edited by a moderator:
Can anyone verify? Apologies for the bump.
 
hi calculusisfun! :smile:

(just got up :zzz: …)
calculusisfun said:
I multiplied d(h) by 16 pi h, because 16 pi h represents the height of an individual slice of the cylinder.

(have a pi: π :wink:)

no, you haven't quite got the hang of this slicing business

the slice is only dh high

(with these slices, the d-something is always one of the dimensions, in this case dh is a length, so you only need the area, you don't need an extra h :wink:)

since its area is πr2, that's a total volume of πr2 dh …

in this case, r is constant, so it's only 16π dh (not 16π h dh)

(btw, i'd keep the 16π*.01 outside a bracket, until just before the end)

try again! :smile:
 
Thanks for the patience tiny-tim. :)

So, are you saying it should actually be:

[PLAIN]http://img820.imageshack.us/img820/3233/eqn3042.png

??

I'm confused. If mass = (density)(volume), why can't I multiply those two formulas together and then integrate??

Sorry for being so uneducated. :p
 
Last edited by a moderator:
hi calculusisfun! :smile:

(just got up :zzz: …)
calculusisfun said:
I'm confused. If mass = (density)(volume), why can't I multiply those two formulas together and then integrate??

you have done …

0.01(8 - h) is the density, and 16πdh is the volume (of each slice)

you need to convince yourself of this!

go through some worked examples from your book (i expect they have things like volume of a sphere from slices … that's with constant density of course) to see it in operation :smile:

(and of course come back here if you're still worried :wink:)
 
Hey Tin-tim,

Again thanks for your help. :)

This question was just on a handout my teacher gave me, so I went looking online for similar problems. I came across this one here.

It looks like they did what I suppose I did. They multiplied the volume function by the density function and then integrated.

It just seems to make sense to me this way haha.

v(h) = 16πh <---- volume of cylinder h high
d(h) = 0.01(8-h) <----- density of cylinder h from the base

So then just multiply them together and integrate that formula from 0 to 8. Why is this incorrect? You said it adds an extra h or something along those lines, but the h in the volume formula and the h in the density formula are independent of each other it appears.

Haha geez I am confused. :p
 
hey calculusisfun! :smile:
calculusisfun said:
… I came across this one here.

It looks like they did what I suppose I did. They multiplied the volume function by the density function and then integrated.

i'm having difficulty viewing that link, but i think there's no integration involved
It just seems to make sense to me this way haha.

v(h) = 16πh <---- volume of cylinder h high
d(h) = 0.01(8-h) <----- density of cylinder h from the base

So then just multiply them together and integrate that formula from 0 to 8. Why is this incorrect? You said it adds an extra h or something along those lines, but the h in the volume formula and the h in the density formula are independent of each other it appears.

ah, when i said "you don't need an extra h", i was talking about dimenions

if you write the area as πr2 (instead of 16π), it becomes clearer …

it should be ∫ (density) πr2 dh, which is ∫ (density) (a volume),

but you're trying ∫ (density) πr2 h dh, which is ∫ (density) (a volume) (a tiny distance)!

tiny distances are still distances! :smile:

to integrate the density times the volume, essentially you do ∫ (density) d(volume), and in this case d(volume) is the volume of the tiny slice
 
  • #10
Thanks for the prompt reply! :)

Oh, so you're saying the volume of a slice is πr^2? But don't you have to take account of the height of the slice? So you would need the h right?
 
  • #11
calculusisfun said:
Thanks for the prompt reply! :)

Oh, so you're saying the volume of a slice is πr^2? But don't you have to take account of the height of the slice? So you would need the h right?

no, the volume of a slice is πr2 dh

dh is a length

(and it's not h because you only want the slice :wink:)

(btw, I'm going out for the evening by half-past)
 

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