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Am I doing this right?

  1. Oct 15, 2004 #1
    ~urgent~ am I doing this right?

    Ok I have these review questions for integration (volumes, applications) but with no answers to them. Could anybody just take a look and say if I got them right? There are NO calculations-- just the integral that represents volume/work/whatever.

    The questions ( 4 of them) are in bold and my answer is after them... :)

    (I'm still not used to text so please bear with me here ^_^;; )

    1- consider a 1 meter rod where the density of the rod varies with the distance x from the left side. let the density be given by g(x) = 1 \ (1 + x^2) kilograms per meter.

    (a) computer the mass of the rod.


    [tex]\int 1 / (1 + x^2) dx [/tex]

    Compute the center of the rod.

    [tex]\int x ( 1/(1 + x^2) ) ds[/tex] over the integral I got in (a).

    limits are both 0 and 1...

    2- Let R be region between the curves y = x^3 and y = x...


    (a) express as an integral the volume of the solid generated by revolving R about the y-axis..


    [tex]\int 3.14 (x - x^3)^2 ds[/tex] limits are 0 and 1

    (b)express the volume of the solid if revolved around the line y = -2...

    [tex]\int 3.14 (-2 - x)^2 - 3.14 (-2 - x^3)^2 dx[/tex] limits 0 and 1

    (c)let S be the solid whose base is R and whose cross-sections perpendicular to x-axis are squares with one side in region R. Express volume of S as an integral...

    If i took a slice here its volume would be delta x times the lenth squared and taking the integral becomes: [tex]\int (x^3 - x)^2 dx[/tex] limits 0 and 1



    3- A cone shaped container sludge is standing upright on its circular end. The diameter of the circular end is 3 meters and the hight of the container is 5 meters and it is filler up to the top with sludge.. the denstiy of sludge at depth of h meter below the surface is given by 1000 + e^(3h^2) kg/m^3...


    (a)how do you break up the barrel into slices that have constant density?


    Since density varies with hight I should take horizontal slices of width w... the radius of the disk at w is (w\2) and the area of the circular disk 3.14(r^) and when multiplied with delta h the volume of each slice is = 3.14 (w\2)^2 delta_h

    (b)what is the work required to pump out a regular slice?


    work = force x distance

    which is the integral of force with respect to h.

    force on each slice is = volume x gravity constant (9.8) x Density

    I'd right the expression but its too messy to type out and I just want to know if I got the above correct since that's the heart of the problem. :blushing:

    (c)Write an integral that gives the total work done to pump all of the sludge over the top rim of the barrel..

    That would be the equation I got above integrated with respect to h...


    4-Set up a definite integral giving the total force due to water pressure on the front face of a fish tank that is 1.5 meter tall and 3 meters side.. recall that the density of water is 1000 kg/m^3 and that g = 9.8 m/s^2...

    well pressure of water is = 9800 (h) where h is hight

    and area of the strip I made (w) in terms of h is w = (4.5-3h)/1.5 delta_h

    its force on that strip alone is water pressure times times w and summing everything up to get an integral gives me =

    [tex[/int9800h (4.5-3h)/1.5 dh[/tex] with limits 0 to 1.5...



    whew.. I have a test in two hours so I guess I should be at the skill level to get all of this right... but who knows. Not doing HW and procrastinating until the last week messes up everything. sigh.... I guess I can only hope that people can change if they want to. :p
     
  2. jcsd
  3. Oct 15, 2004 #2
  4. Oct 15, 2004 #3

    NateTG

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    Homework Helper

    Question 1 looks right to me except that the second integral should be dx rather than ds (probably a typo).

    Question 2:
    You need to be using cylidrical shells, not disks. Parts a and b are definitely incorrect.

    Question 3:
    What I see looks good.

    Question 4:
    I assume the water pressure is in newtons per square meter. Looks good.
     
  5. Oct 15, 2004 #4

    Thanks for the answer!

    About question 2: do you mean I should integrate with respect to y?
     
  6. Oct 15, 2004 #5

    NateTG

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    Too late now, but

    The way that the question is phrased I'd be inclined to go with:
    [tex]\int_0^1 \pi x (x-x^3) dx[/tex]
    Which is the solid of revolution around the y rather than x axis.
     
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