Am I doing this right?

  • Thread starter FancyNut
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  • #1
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~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
 

Answers and Replies

  • #2
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.....

:cry:
 
  • #3
NateTG
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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.
 
  • #4
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NateTG said:
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.

Thanks for the answer!

About question 2: do you mean I should integrate with respect to y?
 
  • #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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