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I need help finding a formula to find the volume of a solid that has a fixed width and depth but 4 different heights. To see what I mean, please take a look at a sketch I drew below.

Thanks.

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- Thread starter beamthegreat
- Start date

- #1

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I need help finding a formula to find the volume of a solid that has a fixed width and depth but 4 different heights. To see what I mean, please take a look at a sketch I drew below.

Thanks.

- #2

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- #3

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- #4

abbas_majidi

- #5

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By shape, I meant how the z-value of the top face changes wrt the domain bounded by the top face.

- #6

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By shape, I meant how the z-value of the top face changes wrt the domain bounded by the top face.

All angles are perpendicular to each other with the exception of the lower vertices. If you cover the bottom of this solid, it would look exactly like a rectangular prism.

Another way to put it is that if the height were all the same, eg.) 1.2, you could calculate the volume by multiplying 2 x 3 x 1.2

- #7

Mark44

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$$\int_{x = 0}^2 \int_{y = 0}^3 ax + by + c~dy~dx$$

- #8

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I've attached an example image of what I mean. The model is the same as the one in your image. Notice the drastic change of color on the top face? That's because the quadrilateral polygon is made up of two triangles. If the polygon was planar, the triangles would be in the same plane and there'd be no change in color. Now I can also choose to retriangularize the polygon so the common edge of the triangles runs between the two other vertices. It would hence look differently and possibly have a different volume. So it's important to know the function z=f(x,y) that describes this top face since it's clearly not planar.All angles are perpendicular to each other with the exception of the lower vertices. If you cover the bottom of this solid, it would look exactly like a rectangular prism.

Another way to put it is that if the height were all the same, eg.) 1.2, you could calculate the volume by multiplying 2 x 3 x 1.2

Now I don't know what you mean by angles being perpendicular to each other. I've checked the angles for each pair of adjacent edges and if the edges are straight then no pair is at a right angle. Check it yourself by calculating the vectors for each edge and taking the dot product.

- #9

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$$\int_{x = 0}^2 \int_{y = 0}^3 ax + by + c~dy~dx$$

Thanks but how do I find the equation of the plane that represents the top face?

- #10

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I've attached an example image of what I mean. The model is the same as the one in your image. Notice the drastic change of color on the top face? That's because the quadrilateral polygon is made up of two triangles. If the polygon was planar, the triangles would be in the same plane and there'd be no change in color. Now I can also choose to retriangularize the polygon so the common edge of the triangles runs between the two other vertices. It would hence look differently and possibly have a different volume. So it's important to know the function z=f(x,y) that describes this top face since it's clearly not planar.

Now I don't know what you mean by angles being perpendicular to each other. I've checked the angles for each pair of adjacent edges and if the edges are straight then no pair is at a right angle. Check it yourself by calculating the vectors for each edge and taking the dot product.

Wow, first off thank you for taking the time to write this response. I haven't really thought it out that far. Basically, what I'm trying to do is to estimate the volume of a pit excavation site by measuring all 4 sides of the pit. How do you suggest I find an estimation of the volume of this pit?

- #11

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It's a surface, but it's not a plane.Thanks but how do I find the equation of the plane that represents the top face?

For any two points in a plane defined by the equation [itex]a x + b y + c z + d = 0[/itex], the equation [itex]a \Delta x + b \Delta y + c\Delta z = 0[/itex] must apply, where [itex]\Delta x[/itex] represents the difference in x-coordinates and so on. This doesn't apply to the top face in question, since two opposite and parallel edges of equal lengths have different [itex]\Delta z[/itex].

- #12

Mark44

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You have the coordinates of all four vertex points on the top face. Label these points A, B, C, and D (where A and C are opposite vertices, and B and D are opposite vertices).Thanks but how do I find the equation of the plane that represents the top face?

Form two vectors, ##\vec{AB}## and ##\vec{AD}##.

Find the cross product of these vectors and call it ##\vec{n}##. This will be a vector that is perpendicular to (normal to) the plane of the top face.

Take any of the four vertices, say ##A = (x_a, y_a, z_a)##.

Let P(x, y, z) be any point on the top plane. Then the equation of the plane is given by: ##\vec{PA} \cdot \vec{n} = 0##.

- #13

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Well if you don't have any information on the interior then the best I think you can do without further information is to fit a plane to those coordinates and hope it doesn't vary too much from reality. Least-squares might do. After that, you use this approximation in the integral given by Mark44.Wow, first off thank you for taking the time to write this response. I haven't really thought it out that far. Basically, what I'm trying to do is to estimate the volume of a pit excavation site by measuring all 4 sides of the pit. How do you suggest I find an estimation of the volume of this pit?

- #14

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Well if you don't have any information on the interior then the best I think you can do without further information is to fit a plane to those coordinates and hope it doesn't vary too much from reality. Least-squares might do. After that, you use this approximation in the integral given by Mark44.

If its not too much to ask, can you quickly do the calculation needed to find the volume of this solid so I can have something to refer to? Thanks!

- #15

Mark44

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Why don't you give it a shot, and we'll take a look at what you did?If its not too much to ask, can you quickly do the calculation needed to find the volume of this solid so I can have something to refer to?

- #16

SteamKing

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There's no explicit formula to calculate the volume of such a solid, and using integral calculus and a hefty dose of analytic geometry seems to be a bit much to figure the volume of a hole.

I need help finding a formula to find the volume of a solid that has a fixed width and depth but 4 different heights. To see what I mean, please take a look at a sketch I drew below.

Thanks.

A quick and dirty estimate of the volume can be made by calculating the areas of the ends of the excavation. These are trapezoids with a width of 2 units each. The heights indicated on your diagram are the lengths of the sides of each trapezoid:

http://www.mathopenref.com/trapezoidarea.html

Let's call these areas A

V ≈ (1/2) * Length * (A

Here, Length = 3 units

If you need something a little more accurate, you can calculate the area of a section of the excavation in the middle and use Simpson's First Rule, but the trapezoidal rule should give you a sufficiently accurate result for most purposes.

- #17

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There's no explicit formula to calculate the volume of such a solid, and using integral calculus and a hefty dose of analytic geometry seems to be a bit much to figure the volume of a hole.

Thank you, SteamKing! This is exactly what I needed.

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##A(x) = \frac{1}{2}(2)(h_1 + h_2) = h_1 + h_2##

You could integrate this for x from 0 to 3, or (as it's linear) just take the averages of the h's and multiply by 3.

This gives ##V = 3(1.35 + 1.35) = 8.1##

Am I missing something?

- #19

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I thought it was worth posting the general solution here. If the solid has width ##w##, length ##l##, heights ##a, b## at the front corners and heights ##c, d## at the back, then:

##h(x, y) = a + \frac{y}{w}(b-a) + \frac{x}{l}[c + \frac{y}{w}(d - c) - (a + \frac{y}{w}(b-a))]##

Where ##x## is the distance along the length and ##y## across the width and ##h(x,y)## is the height at that point.

##\int_{0}^{l} \int_{0}^{w}h(x, y)dydx = wl(\frac{a+b+c+d}{4})##

So, the volume is ##wl## times the average height of the four corners, which makes sense.

##h(x, y) = a + \frac{y}{w}(b-a) + \frac{x}{l}[c + \frac{y}{w}(d - c) - (a + \frac{y}{w}(b-a))]##

Where ##x## is the distance along the length and ##y## across the width and ##h(x,y)## is the height at that point.

##\int_{0}^{l} \int_{0}^{w}h(x, y)dydx = wl(\frac{a+b+c+d}{4})##

So, the volume is ##wl## times the average height of the four corners, which makes sense.

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- #20

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This is of course assuming the surface is a hyperbolic paraboloid, which is not necessarily the case. But it works as an approximation.I thought it was worth posting the general solution here. If the solid has width ##w##, length ##l##, heights ##a, b## at the front corners and heights ##c, d## at the back, then:

##h(x, y) = a + \frac{y}{w}(b-a) + \frac{x}{l}[c + \frac{y}{w}(d - c) - (a + \frac{y}{w}(b-a))]##

Where ##x## is the distance along the length and ##y## across the width and ##h(x,y)## is the height at that point.

##\int_{0}^{l} \int_{0}^{w}h(x, y)dydx = wl(\frac{a+b+c+d}{4})##

So, the volume is ##wl## times the average height of the four corners, which makes sense.

An alternative approximation is to use two diagonally opposite points at heights [itex]z_1, z_2[/itex]. This allows you to calculate the volume as [itex]V = \frac{z_1 + z_2}{2} \Delta x \Delta y[/itex] by interpreting the surface as a plane.

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- #21

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This is of course assuming the surface is a hyperbolic paraboloid, which is not necessarily the case. But it works as an approximation.

An alternative approximation is to use two diagonally opposite points at heights [itex]z_1, z_2[/itex]. This allows you to calculate the volume as [itex]V = \frac{z_1 + z_2}{2} \Delta x \Delta y[/itex] by interpreting the surface as a plane.

The surface is not a plane, but every cross-section (parallel to the sides) is a trapezium. The formula I gave for ##h(x,y)## is exact at every point.

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