Am I doing this right?

1. Oct 15, 2004

FancyNut

~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.

$$\int 1 / (1 + x^2) dx$$

Compute the center of the rod.

$$\int x ( 1/(1 + x^2) ) ds$$ 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..

$$\int 3.14 (x - x^3)^2 ds$$ limits are 0 and 1

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

$$\int 3.14 (-2 - x)^2 - 3.14 (-2 - x^3)^2 dx$$ 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: $$\int (x^3 - x)^2 dx$$ 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.

(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. Oct 15, 2004

FancyNut

.....

3. Oct 15, 2004

NateTG

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. Oct 15, 2004

FancyNut

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

5. Oct 15, 2004

NateTG

Too late now, but

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