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Calc book question

  1. Apr 6, 2005 #1
    question is here
    answer for b is here

    My question is, why did the book use [tex]4\pi x^2 \Delta x[/tex] as the formula for volume? isn't that the derivative of the volume? since mass=density*volume, shouldn't the equation for volume be [tex]\frac{4}{3} \pi r^3[/tex]
     
  2. jcsd
  3. Apr 6, 2005 #2
    That is the derivative of volume, correct. They integrate the derivative multiplied by the density function to get the mass. You can conceptualize it as adding up the masses of infinitely many infinitessimally thin shells, going from radius 0 to radius 6.

    The volume of each infinitessimally thin shell is its area multiplied by the thickness [itex]dx[/itex]. Multiply this volume by the density function to get the integrand.
     
  4. Apr 6, 2005 #3
    why does the book integrate the density?

    you never said "They integrate the derivative multiplied by the integral of density to get the mass."
     
  5. Apr 6, 2005 #4
    I should have been more mathematical and said "they integrate [the derivative multiplied by the density function]" :smile:

    Think of it this way: The area of a certain shell at radius [itex]r[/itex] from the center is [itex]4\pi r^2[/itex], so the volume of a very thin shell at radius [itex]r[/itex] from the center is approximately [itex]4 \pi r^2 dr[/itex] where [itex]dr[/itex] is the thickness (as the thickness goes to zero, the approximation becomes better and better, and in the limit it is exact). The mass of this shell is then given by its volume times its density, and the density at radius [itex]r[/itex] is given as

    [tex]\frac{1}{1+\sqrt{r}}[/tex]

    so the mass of such a shell is

    [tex]\frac{4\pi r^2}{1+\sqrt{r}} \ dr,[/tex]

    and to get the mass of the whole ball, we add together the masses of all such shells from radius zero to radius [itex]6[/itex], ie. take

    [tex]\int_0^6 \frac{4\pi r^2}{1+\sqrt{r}} \ dr.[/tex]
     
    Last edited: Apr 6, 2005
  6. Apr 6, 2005 #5
    sooo... it's more like "they integrate [the surface area multiplied by the density function]" ?
     
  7. Apr 6, 2005 #6
    Yeah. Read my argument below that and see if you understand why they'd be doing that :smile:
     
  8. Apr 6, 2005 #7
    yes, I understand totally, I just didnt know why the bok would write volume when it obiously is surface area. That just go me cofused
     
  9. Apr 6, 2005 #8
    good :smile:
     
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