Calculating Roadbed Volume

[tatertoph]

Homework Statement

Determine the number of cubic yards of crushed rock necessary to make a roadbed with the cross-section shown below, if the road is to be 1/4 mile − long. Assume that the crown of the pavement is an arc of a parabola, with its vertex at the top center. Be careful with units.

The Attempt at a Solution

First i found the volume of the lower part of the road bed and converted inches to feet. V= l*w*h =(1320ft)(20ft)(2/3ft) = 17600 ft^3 . 17600ft/3 = 5866.667 yds^3.

Now I have to find the top part of the road which is where I'm stuck.
I got an integral that was V= 1320ft * "integral from 0 to 4" (-x^2-x+4)dx which is
V= 1320 ft * ((-x^3/3)-(x^2/4)+4x).
Then I plugged in 4 and got -17600 ft^3.

That can't be right though because when you add those volumes you get 0. What did I do wrong?

 

[Spinnor]

Is the x-axis horizontal in your picture?

 

[tatertoph]

Is the x-axis horizontal in your picture?

yes

 

[Spinnor]

yes

I think your integral is set up wrong? Your x integral should run 20 feet?

 

[Mark44]

Where is the horizontal axis in your drawing? You can put it anywhere, but where you put it affects the equation of your parabola.

 

[tatertoph]

Where is the horizontal axis in your drawing? You can put it anywhere, but where you put it affects the equation of your parabola.

I think I figured it out. I forgot to convert the 4 inches to feet. So instead of integrating from 0 to 4 it would be 0 to 1/3. Then my final answer would be 1670.3 feet^3 or 556.8 yd^3.

As you can see I am integrating squares (ΔA) from the base to the apex of the curve.

I'm not sure if this is right still and I would appreciate it if someone could do the problem out and show me how they did it.

Edit: I messed up drawing that! In the picture the axis of symmetry is supposed to go down the center of the shape, not down the side.

 

[tatertoph]

Can anyone help me out?

 

[Spinnor]

Can anyone help me out?

Please show your integral for the second part. I still think something is off.

 

[Nidum]

Just for interest - the engineers solution :

The top curve is very flat - the local radius at any point is very large . Given this and the fact that the exact equation for the parabolic shape that you are told to assume is not given then we can reasonably use a plain arc .

If top curve is an arc then area of top part of the cross section is just area of a segment .

 

[Spinnor]

is not given

It is not given but can be found from the dimensions of the drawing? It has a width of 20 feet and a height of 1/3 foot, only one parabola fits?

y = ax^2 + bx + c

(-10,0), (10,0), and (0,1/3) are the set of points of the parabola in one reference system.

 

[haruspex]

Then my final answer would be 1670.3 feet^3 or 556.8 yd^3.

I strongly recommend doing simple calculations to find the ballpark first.
The road cross section is not far short of being one foot all the way across. That gives 20 sq ft area, times 1320 ft = 26400 cu ft. So your 1670 is way short.
You also converted to cu yd wrongly. How many cu ft in a cu yd?
For the accurate answer, the easiest way is to start with the 26400 and calculate how much to take off. For that, you can measure y downwards from the one foot mark.

 

[tatertoph]

I made a lot of mistakes when I first posted this question but I fixed them and handed it in.
I got 5 points off on this question. Looks like i messed up the equation of the parabola, but I figured I would post this if someone else needs it.

 

[haruspex]

I made a lot of mistakes when I first posted this question but I fixed them and handed it in.
I got 5 points off on this question. Looks like i messed up the equation of the parabola, but I figured I would post this if someone else needs it.View attachment 113608 View attachment 113607

I'm not surprised you lost some marks. Your images are very dark and hard to read, but the last line seems to say over 6000 yd³. Did you not read/understand my ballpark calculation?
The cross sectional area is a bit less than 20ft². The length is 1320ft. That gives an upper limit of 26400 ft³. How many cubic feet in a cubic yard? (Hint: you should get less than 1000yd³.)