# Volume Via Slices

1. May 26, 2014

### Shinaolord

1. The problem statement, all variables and given/known data

A dam has a rectangular base 1500 meters long and 140 meters wide. Its cross-section is shown in the figure. (The Grand Coulee Dam in Washington state is roughly this size.) By slicing horizontally, set up and evaluate a definite integral giving the volume of material used to build this dam.

The height at the top is 10m, bottom is 140m, and the height is 130m.

2. Relevant equations
I tried to use similar triangles but I really have no idea how to do this.
Any suggestions?

3. The attempt at a solution
130-h / 130 = w/140
w= 130-(h/130)
but integrating this and multiplying by the base (1500) gives the wrong answer.

2. May 26, 2014

### SammyS

Staff Emeritus
Can you post the figure?

3. May 26, 2014

### Shinaolord

Yes, here.

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4. May 26, 2014

### LCKurtz

When h= 0 you are getting w = 130 and when h = 130 you are getting w = 129. Both wrong. Redo your linear equation. You will know it's right when you get the correct values for w.

5. May 27, 2014

### Shinaolord

So, I am not even sure how to obtain the formula, i tried using the two points (5,130) and (70,0). I got this by placing the coordinate system's origin down the middle of the dam. Doing this I got the equation

$$f(x) = -12x +190$$
maybe insead of 12, the slope i 1.2?

which is obviously wrong. But I want to explain my method so I know if I'm at least on the right path

I defined A(x) as the area between y=130 and f[x], giving me the formula
/begin{equation*}$$\int_{0}^{5} 130 dx + \int_{5}^{70} f(x) dx =\frac{A(x)}{2}$$/end{equation*}
The 1/2 A(x) comes from the symmetry across the y axis using my coordinate system.
Is that part right? FOr some reason, finding the equation of the line is proving to be very difficult even though I know how to find the formulas for lines. Geometric problems are challenging for me.

Last edited: May 27, 2014
6. May 27, 2014

### Shinaolord

I have an idea. Would the slope, m, be equal to -130/70?

it seems correct.

7. May 27, 2014

### SammyS

Staff Emeritus
If you are using the points, (5,130) and (70,0) , then the slope is not -130/70 , although that's pretty close.

Neither is it -12 nor -1.2 .

Slope is rise over run, right?

8. May 27, 2014

### Shinaolord

-130/65?

Wow, I feel stupid.

9. May 27, 2014

### SammyS

Staff Emeritus
Yup.

and -130/65 = -2 .

10. May 27, 2014

### Shinaolord

So, I can just use that equation as F of X in the integral equation I posted above? And if not where did I go wrong in that equation? Please don't explicitly tell me, but hints are great thank you very much for your assistance

Edit:
If this post in the above post look weird, it's because I'm using Siri on my iPhone to translate my words into text and it's not good with math

11. May 27, 2014

### SammyS

Staff Emeritus
What do you now get for f(x) ?

12. May 27, 2014

### Shinaolord

I get $$f(x) = -2x +140$$.

13. May 27, 2014

### SammyS

Staff Emeritus
Looks good.

Now set-up that integral.

14. May 27, 2014

### Shinaolord

Okay, so...

$$\int_{0}^{5} 130 dζ + \int_{5}^{70} -2ζ+140 dζ = \frac{A(ζ)}{2}$$

Which come out to
$$\int_{0}^{5} 130 dζ = 650$$
$$\int_{5}^{70} -2ζ+140 = 4,255$$

$$A(ζ) = 2( 650+4,255) = 9,750$$ m2

Now, we have a dam that is 1,500 meters long, and

$$V(ζ , x ) = A(ζ) *Δx$$ where $$Δx = 1,500$$ m

$$V(ζ , x ) = 9,750*1,500 = 14,625,000$$ m3

15. May 27, 2014

### LCKurtz

That is correct. Note that this problem could have been done without calculus. Using the midline of the trapezoid whose length is the average of the top and bottom widths you get (ave.width)(height)(depth) = (75)(130)(1500) =14,625,000

16. May 28, 2014

### Shinaolord

Yes I realize that and I used that to check my answer at the end, but were supposed to use slices to find it otherwise we do not get points so I had to do it this way thanks for the alternative method though!