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Linear approximation of paint

  1. Mar 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Use linear approximation to estimate the amount of paint in cubic centimeters needed to apply a coat of paint 0.100000 cm thick to a hemispherical dome with a diameter of 45.000 meters.

    2. Relevant equations
    [tex]Surface Area of sphere=4\pi(r^2)[/tex]
    Since it is hemipshereical, the surface area will be half
    [tex]Surface Area of hemispherical dome=2\pi(r^2)[/tex]
    [tex]dSA=4\pi(r)dr[/tex]


    3. The attempt at a solution
    I converted 45m into 4500cm for the radius. I set dr=.1cm
    and the radius to 4500cm.
     
    Last edited: Mar 9, 2007
  2. jcsd
  3. Mar 9, 2007 #2

    nrqed

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    The question says that 45 m is the diameter, not the radius.
     
  4. Mar 9, 2007 #3
    Oops. Well I inputed 2250cm for the radius and it is still wrong
     
    Last edited: Mar 9, 2007
  5. Mar 9, 2007 #4
    Am I approaching this the right way?
     
  6. Mar 9, 2007 #5

    Dick

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    This time the problem is to determine a volume. So you want to estimate the change in volume if a hemisphere grows from diameter 45m to 45.002m.
     
  7. Mar 9, 2007 #6

    HallsofIvy

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    As you have been told the DIAMETER is 45 m. so the radius is 22.5 m= 2250 cm. In addition, YOU said
    but then say
    Shouldn't it be
    [tex]dSA= 2\pi r^2 dr[/tex]?
     
  8. Mar 9, 2007 #7

    HallsofIvy

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    As you have been told the DIAMETER is 45 m. so the radius is 22.5 m= 2250 cm. In addition, YOU said
    but then say
    You don't want Surface area, you want VOLUME. The volume of a sphere is [itex]\frac{4}{3}\pi r^3[/itex]. The differential is [itex]dV= \frac{4}r^2 dr[/itex] which is exactly the same as the surface area times the "thickness" dr. I thought that was what you were doing when you quoted the formula for surface area!
     
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