Assignment 9

course Mth 174

19:25:37query problem 8.4.3 (8.3.4 3d edition) was 8.2.6) moment of 2 meter rod with density `rho(x) = 2 + 6x g/m

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A.) sum(2+6xsub i)dx.

B.)Int.(2+6x,0,2)dx = 3x^2+2x = 3(4)+2(2)-0 = 16

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20:15:30

what is the moment of the rod?

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Int.(x(2+6x),0,2)dx = 2^2+2(2)^3 = 4+16 = 20

Very good. See the following for more detail and some additional information that shouldn't be difficult to understand having correctly solved the stated problem.

The mass of an increment of length `dx, with sample point x_i, is (2 + 6 x_i ) `dx.

The moment is mass * distance from axis of rotation. Assuming axis of rotation x = 0:

The moment of the mass in the increment is (2 + 6 x_i) * x_i.

Setting up a Riemann sum and allowing the increment to approach zero, we obtain the integral

moment = int(x(2+6x), x, 0, 2).

Thus the integrand is 2x + 6 x^2. An antiderivative is F(x) = x^2 + 2 x^3, so the definite integral is

moment = int(x(2+6x), x, 0, 2) = F(2) - F(0) = 20.

The moment of the typical increment has units of mass/unit length * length * distance from axis, or (g / m) * m * m = g * m.

The units of the integral are therefore g * m, and the moment of this object is 20 g * m (i.e., 20 gram * meters).

ADDITIONAL INFORMATION (finding center of mass):

To get the center of mass relative to x = 0 (this was not requested here but you should know how to do this), divide the moment about x = 0 by the mass of the object:

The mass of the object is easily found to be int((2+6x), x, 0, 2) (see above for the mass increment, which leads to the Riemann sum then to the integral).

The center of mass is therefore

center of mass = moment / mass = int(x(2+6x), x, 0, 2) / int((2+6x), x, 0, 2).

The integrand for the denominator is 2 + 6 x, antiderivative G(x) = 2x + 3 x^2 and definite integral G(2) - G(0) = 16 - 0 = 16, meaning that the object has mass 16 grams (more specifcally the units of the denominator will be units of mass / unit length * length = mass, or in this case g / m * m = g).

So the center of mass is at x = 20 g * m /(16 g) = 5/4 (g * m) / g = 5/4 g. **

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20:15:54

What integral did you evaluate to get a moment?

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RESPONSE -->

Int. (x(2+6x),0,2)

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20:55:11

query problem 8.4.12 (was 8.2.12) mass between graph of f(x) and g(x), f > g, density `rho(x)

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Int. (f(rho(x))-g(rho(x)))dx

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20:55:44

what is the total mass of the region?

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RESPONSE -->

Int. (f(rho(x))-g(rho(x)))dx

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20:55:57

What integral did you evaluate to obtain this mass?

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RESPONSE -->

Int. (f(rho(x))-g(rho(x)))dx

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21:21:14

What is the mass of an increment at x coordinate x with width `dx?

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Int.(rho(x),a,x)dx

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21:22:40

What is the area of the increment, and how do we obtain the expression for the mass from this area?

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area = width(length) = xdx. Expression for mass is the integral of xdx(density).

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21:34:03

How to we use the mass of the increment to obtain the integral for the total mass?

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We use an integral to add the masses of all of the similar increments.

You basically got the problem, but some of your subsequent answers were questionable, so be sure you check out the following:

First you find the mass of a typical increment of width `dx, with sample point x within the interval.

The mass is just area * density.

The area of the region is height * width, or approximately (f(x) - g(x) ) * `dx. The density is `rho(x) so you get the approximation

`dm = area * density
= (f(x) - g(x) ) * 'dx * `rho(x)
= `rho(x) (f(x) - g(x) ) * 'dx.

The Riemann sum is the sum of all such mass increments for a partition of the interval, and at the interval width `dx approaches 0 this sum approaches the integral

int( rho(x) * (f(x) - g(x)), x, a, b).

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21:36:19

query problem 8.5.13 (8.4.10 3d edition; was 8.3.6) cylinder 20 ft high rad 6 ft full of water

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23:04:23

how much work is required to pump all the water to a height of 10 ft?

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8.49 ft-lbs.

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23:05:11

What integral did you evaluate to determine this work?

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Int. (90pi(9.8)h^2(20-h),0,10) and then converted from Joules into Foot Pounds.

Overall a good solution.

However check the following for a couple of discrpancies (e.g., the limits on the integration and the height to which water is raised):

** For the ith interval, using F_i and dist_i for the force required to lift the water and the distance is must be lifted, `rho for the density and A for the cross-sectional area 36 `pi, the work is

Fi * dist_i = `rho g A 'dyi * (30 - y_i) = `rho g A (30 - yi) 'dy_i.

We thus have a Riemann sum of terms `rho g A (30 - y_i) 'dy_i.

This sum approaches the integral

int(`rho g A (30 - y) dy between y = 0 and y = 20).

The limits are y = 0 and y = 20 because that's where the fluid represented by the terms `rho g A (30 - y_i) 'dyi is (note that the tank for Problem 6 is full of water, not half full). The 30-y_i is because it's getting pumped to height 30 ft.

Your antiderivative is `rho g A ( 30 y - y^2 / 2).

At this point you can plug in your values for `rho, g and A, then evaluate 30 y - y^2 / 2 at the limits to get your answer.

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23:13:30

Approximately how much work is required to pump the water in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.

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Work = Force x Distance = (density x gravity x Vol.)(y-dy)

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23:14:40

Explain how your answer to the previous question leads to your integral.

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You must use the sum of every small strip to find the answer, therefore you use an integral of the work done on one strip from the lower limit to the upper limit.

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00:01:46

query problem 8.3.18 CHANGE TO 8.5.15 work to empty glass (ht 15 cm from apex of cone 10 cm high, top width 10 cm)

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00:08:56

how much work is required to raise all the drink to a height of 15 cm?

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RESPONSE -->

1.44x10^7 Joules

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00:09:29

What integral did you evaluate to determine this work?

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RESPONSE -->

Int.(111.111pih^2(9.8)(15-h),0,15)

Not bad. Compare with the following:

** The diameter of the top of the cone is equal to the vertical distance y from the apex to the top.

At height y the diameter of the cone is easily seen to be equal to y, so the cross-section at height y has radius .5 y and therefore area A = `pi ( .5 y ) ^ 2 = `pi / 4 * y^2.

A slice of thickness `dy at height y has approximate volume A * `dy = `pi/4 * y^2 * `dy.

This area is in cm^3 so its mass is equal to the volume and the weight in dynes is 980 * mass = 245 `pi y^2 `dy.

This weight is raised from height y to height 15, a distance of 15 - y. So the work to raise the slice is force * distance = 245 `pi y^2 `dy * ( 15 - y ) = 245 ( 15 - y ) `pi y^2 `dy = 245 `pi ( 15 y^2 - y^3 ) `pi `dy

Slices go from y = 0 to y = 10 cm so the integral is 245 `pi ( 15 y^2 - y^3 ) `pi dy, evaluated from y = 0 to y = 10.

We get 245 `pi * (5 y^3 - y^4 / 4) evaluated between 0 and 10.

The result is 245 `pi * 2500 ergs, close to 2 million ergs.

Most calculations were mentally so check the precise numbers. The process is correct. **

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00:15:04

Approximately how much work is required to raise the drink in a slice of thickness `dy near y coordinate y? Describe where y = 0 in relation to the tank.

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About 1.44*10^7

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00:15:39

How much drink is contained in the slice described above?

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Hardly any at all.

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00:22:11

What are the cross-sectional area and volume of the slice?

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Vol. = pi*r^2*dy

Area = pi*r^2

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00:22:46

Explain how your answer to the previous questions lead to your integral.

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You must use the integral of the volume in conjunction with the upper and lower limits in order to find the total volume.

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00:22:55

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Assignment 9

Overall your work looks good, though it doesn't always agree with the solutions I've inserted.

&#See my notes and let me know if you have questions. &#