my sloping rear yard - [imath]\bold{Not}[/imath] my thread

[logistic_guy]

This post was sent to me by the new member @dw1803.

Anyone know calculus?

I need an answer to this math problem.

Hi, I need help with figuring out what volume I have for my sloping rear yard.

The area is a rectangular 12’ x 18’

The shallow end starts at ground level.

One end of the 12’ length falls 8”

The other end of the 12’ length falls 13”

What would be the volume in cubic yards?

 

[The Highlander]

This post was sent to me by the new member @dw1803.

Anyone know calculus?

I need an answer to this math problem.

Hi, I need help with figuring out what volume I have for my sloping rear yard.

The area is a rectangular 12’ x 18’

The shallow end starts at ground level.

One end of the 12’ length falls 8”

The other end of the 12’ length falls 13”

What would be the volume in cubic yards?

I'm having a great deal of difficulty trying to even visualize this situation (given the way it has been described). The best I could come up with would be something like this...

​

I suppose that might be described as a (summable?) series of right angled triangles whose bases remain constant (at 12') but whose heights vary from 8' to 13' along a length (axis?) of 18'.

To remind you of the Geometry (not your strong point you repeatedly profess), the area of a triangle is given by: half the base multiplied by the height, so the area of the "front" face would be: ½ × 12 × 8 = 48 ft² whilst that of the far (rear) face would (similarly) be: 78 ft².

I expect the rate at which the height is increasing would be given by the slope of the topmost line in the diagram, ie: \(\displaystyle \frac{5}{18}\) (feet per foot?).

You are the "expert" in Calculus. How would you go about summing up these areas given the parameters I have suggested above?

You should also start a conversation with @dw1803 so that s/he may submit anything s/he may wish to contribute to the thread now that you have started it on their behalf. (Like, for example, a clearer description of the situation; I don't understand exactly how if the "shallow end starts at ground level" then the yard "falls" away from there. )

PS: A cubic yard is 27 cubic feet so, if you can calculate the volume in ft³ then you only need divide by 27 to get the answer in the units requested.

Hope that helps.

PPS: I would guess: \(\displaystyle \int_{8}^{13}6h.dh\;\) (since, if h is the (varying) height, 6h gives the area of each triangle in ft²) but that may just be rubbish! What would you (@logistic_guy) advise?

 

[khansaheb]

This post was sent to me by the new member @dw1803.

Anyone know calculus?

I need an answer to this math problem.

Hi, I need help with figuring out what volume I have for my sloping rear yard.

The area is a rectangular 12’ x 18’

The shallow end starts at ground level.

One end of the 12’ length falls 8”

The other end of the 12’ length falls 13”

What would be the volume in cubic yards?

@dw1803

Do you have an approximate sketch depicting the yard in 3-D?

If you do - please post the sketch. A photograph, using your cell-phone will be valuable. The problem statement is ambiguous. You can reply to this message to post.

 

[The Highlander]

@dw1803

Do you have an approximate sketch depicting the yard in 3-D?

If you do - please post the sketch. A photograph, using your cell-phone will be valuable. The problem statement is ambiguous. You can reply to this message to post.

Are you sure that new (as yet, 'unapproved') members can post a (visible) "reply" in a thread?

I wouldn't have expected that to be any more possible than them being able to post in a new thread but, if you've seen that done or know it's possible, that would be really handy! Can you confirm it is actually possible?

That's why I suggested that @logistic_guy should open a conversation with @dw1803 (since he started this thread for him or her) to give them the "facility" to make (further) contributions.

 

[khansaheb]

Are you sure that new (as yet, 'unapproved') members can post a (visible) "reply" in a thread?

No ... I am not sure. Probably your solution is more feasible.

 

[The Highlander]

Postscript: My "guess" \(\displaystyle \left(\int_{8}^{13}6h.dh\right)\;\) can't possibly be right; it only produces 315 ft³!

Clearly, it failed to take into account the integration over the 18' width of the yard. Oh well!

(Could 210 (or 36) cubic yards be the right answer? Hmmm, those seem like awfully neat results for a "real-life" problem. )

 

[logistic_guy]

What would you (@logistic_guy) advise?

If your diagram is correct, the volume can be calculated with two integrals.

You are the "expert" in Calculus. How would you go about summing up these areas given the parameters I have suggested above?

I still did not read the OP problem. I was giving a chance to others to contribute. In other words, I am giving khan the opportunity to be useful one time and help the OP. khan is being very useless during the \(\displaystyle 2025, 99\%\) of his \(\displaystyle 1000\) posts are just rubbish.

Also the OP did not try a calculus attempt, so it's boring to guess what he exactly needs!

You should also start a conversation with @dw1803 so that s/he may submit anything s/he may wish to contribute to the thread now that you have started it on their behalf. (Like, for example, a clearer description of the situation; I don't understand exactly how if the "shallow end starts at ground level" then the yard "falls" away from there. )

Sure I will. My message board is open to any members. If the OP reads this and he is serious, I am sure that he will message me to discuss.

Are you sure that new (as yet, 'unapproved') members can post a (visible) "reply" in a thread?

I am not sure if new members can message us. But I was the one who started the conversation to present Help.

That's why I suggested that @logistic_guy should open a conversation with @dw1803 (since he started this thread for him or her) to give them the "facility" to make (further) contributions.

I am opening. And I am always ready for the OP conversation.

Sir @The Highlander I was having difficulties to reply to your posts because your font is very big. lol

 

[The Highlander]

Final thought (from me):-

I've just worked out that, if the height of the triangle remained constant (@ 8') along the 18' length, then that would be a triangular prism whose volume was 864 ft³, whilst, if the height of the triangle remained constant (@ 13') along the 18' length, then that would also be a triangular prism but its volume would be: 1,404 ft³.

Since, the rate of increase in the height of the triangle(s) appears to be constant along the 18' width of the yard, then I would expect the mean value of those two prisms' volumes might be a valid approximation of the actual volume involved here, ie: 1,134 ft³ (or 42 cubic yards).

Though I expect a properly constructed definite integral would arrive at a more exact answer?

 

[The Highlander]

If your diagram is correct, the volume can be calculated with two integrals.

Which are? (I'd be interested to see.)

Sir @The Highlander I was having difficulties to reply to your posts because your font is very big. lol

If you hit "Reply" to any of my posts and find the size of the quoted text inconvenient, simply select everything in the quoted part and choose a smaller text size (eg: 18 or even 15).

For example, notice how I have changed the size of the text in your quoted message (above). Simples!

 

[logistic_guy]

Which are? (I'd be interested to see.)

One volume for the triangular prism with height \(\displaystyle 8\) which can be calculated directly without calculus. And the other volume is just the rest of the volume of height \(\displaystyle 5\). I will give a diagram for it.



I guess it can be calculated without calculus (not sure). But it has a beautiful formula and it's easy to be set with an integral.

For example, notice how I have changed the size of the text in your quoted message (above). Simples!

lol smart but tedious

 

[The Highlander]

One volume for the triangular prism with height \(\displaystyle 8\) which can be calculated directly without calculus. And the other volume is just the rest of the volume of height \(\displaystyle 5\). I will give a diagram for it.

I'm afraid the plane you have cut through this "solid" (in your diagram) doesn't create a prismatic solid on either the upper or lower sections, so calculating the volume of either of these pieces is no more straightforward that calculating that of the whole.

I guess it can be calculated without calculus (not sure). But it has a beautiful formula and it's easy to be set with an integral.

I suspect my approach (at Post #8, qv) may be the closest to finding an (approximate?) answer without the use of Calculus (though I'll be happy to be proved wrong about that) but it was the integral solution that I really wanted to see.
Does it require the use of a double integration \(\displaystyle \left(\underset{\mathbb{R}}{\iint_{a}^{b}}f(x,y)\text{ or similar?}\right)\) and can you (or anyone else) show me what it should be ????

 

[logistic_guy]

One volume for the triangular prism with height \(\displaystyle 8\) which can be calculated directly without calculus. And the other volume is just the rest of the volume of height \(\displaystyle 5\). I will give a diagram for it.

My interpretation seems wrong because I assumed the heights of the two sides are \(\displaystyle 8\) & \(\displaystyle 8\) not \(\displaystyle 8\) & \(\displaystyle 13\)

But I came up with an idea that uses parametrization. I got this result:

\(\displaystyle \int_{0}^{18}6\left(8 + \frac{x}{18}\right) \ dx = 918 \ \text{unit}^3\)

 

[The Highlander]

But I came up with an idea that uses parametrization. I got this result:

\(\displaystyle \int_{0}^{18}6\left(8 + \frac{x}{18}\right) \ dx = 918 \ \text{unit}^3\)

Thank you for that.

I had assumed it might require the use of double integration (as I'd seen discussed here) but I'll take your word for it that what you've shown does the job. It does produce a result (34 cubic yards) in nearly the same 'ball-park' as I got with my geometric approximation so I guess I'm happy.

 

[logistic_guy]

Thank you for that.

I had assumed it might require the use of double integration (as I'd seen discussed here) but I'll take your word for it that what you've shown does the job. It does produce a result (34 cubic yards) in nearly the same 'ball-park' as I got with my geometric approximation so I guess I'm happy.

Here is my approach.

Look at my diagram. I reversed the shape so that I have positive values. The volume will not be affected.

My goal is to represent the height which is shown with a formula. If I move in space from the point \(\displaystyle (0,0,8)\) to \(\displaystyle (18,0,13)\) I can parametrize this segment by \(\displaystyle r(t) = (0,0,8) + t(18 - 0, 0 - 0, 13 - 8) = (0,0,8) + (18t,0,5t) = (18t,0,8 + 5t)\). Thanks to the differential geometry!

This gives:

\(\displaystyle x(t) = 18t\)
\(\displaystyle y(t) = 0\)
\(\displaystyle z(t) = 8 + 5t\)

Therefore, the height of this prism's face (along the \(\displaystyle x\)-axis) is always \(\displaystyle z(t) = 8 + 5t\). Now we will take a slice of the volume of this prism (the pink volume) and increase it to the full \(\displaystyle y\)-axis, the face along the \(\displaystyle y\)-axis is just a triangle. So, the volume of this new slice (increased slice) is just \(\displaystyle \frac{1}{2}12z \ dx\).

We can write the volume of the prism with triple integrals like this:

\(\displaystyle V = \int dV = \frac{1}{2}\int_{0}^{18}\int_{0}^{12}\int_{0}^{z} dz \ dy \ dx\)

Or directly like this:

\(\displaystyle V = \frac{1}{2}\int_{0}^{18} 12z \ dx = \int_{0}^{18}6(8 + 5t) \ dx\)

We know that \(\displaystyle x = 18t\), then \(\displaystyle t = \frac{x}{18}\).

This gives:

\(\displaystyle V = 6\int_{0}^{18}\left(8 + \frac{5x}{18}\right) \ dx = 1134 \ \text{unit}^3\)

I got a different result this time because in the last post I wrote \(\displaystyle 8 + \frac{5x}{18}\) as \(\displaystyle 8 + \frac{x}{18}\), forgetting to multiply \(\displaystyle x\) by \(\displaystyle 5\).

 

[logistic_guy]

My integral can also derive the general formula for any prism like the diagram you provided but with different dimensions.

\(\displaystyle V = \frac{w}{2}\int_{0}^{l}\left(h_2 + \frac{h_1 - h_2}{l}x\right) \ dx = \frac{1}{4}(h_1 + h_2)lw\)

where

\(\displaystyle l\) - length of the base
\(\displaystyle w\) - width of the base
\(\displaystyle h_1\) - longer height
\(\displaystyle h_2\) - shorter height

Using the formula to find the same volume.

\(\displaystyle l = 18\)
\(\displaystyle w = 12\)
\(\displaystyle h_1 = 13\)
\(\displaystyle h_2 = 8\)

\(\displaystyle V = \frac{1}{4}(h_1 + h_2)lw = \frac{1}{4}(13 + 8)18\times 12 = 1134 \ \text{unit}^3\)

I didn't look it up but I think that this formula already does exist.

 

[The Highlander]

\(\displaystyle V = \frac{w}{2}\int_{0}^{l}\left(h_2 + \frac{h_1 - h_2}{l}x\right) \ dx = \frac{1}{4}(h_1 + h_2)lw\)
\(\displaystyle V = \frac{1}{4}(h_1 + h_2)lw = \frac{1}{4}(13 + 8)18\times 12 = 1134 \ \text{ft}^3\) = 42 cubic yards!

Amazing!

So adjusting your integral (I thought it would require multiple integration) to get an even more accurate result actually produces exactly the same answer as my (purely) Geometric interpretation (at Post #8, qv).

I guess Calculus may not have been (absolutely) necessary after all but it was interesting to see your approach and thank you for taking the trouble to show in detail how you arrived at your results.

I expect @dw1803 now has (all?) the information s/he was looking for so it's unlikely we'll ever see or hear from him/her again (not even to offer any thanks for the efforts expended on his/her behalf ). Unless s/he suddenly re-appears with a different diagram or clearer explanation of the problem rendering all of the foregoing a waste of time! lmao.

(Of course, time spent in recce is never a waste of time, as my old CO used to constantly remind us, and this has proved (IMNSHO, lol) an interesting little exercise with some useful results! )

 

[The Highlander]

Afterthought:-

It occurred to me that (some) folks might be interested (or find it useful?) to see how my diagram of the volume was created in Desmos.

The input (to create it) can be viewed here but here's a pretty picture of the result for your enjoyment...

​

Hope that's of interest.

 

[logistic_guy]

Amazing!

Your approach was in fact more amazing. You could deal with this complex prism like it was nothing. I was very surprised when my result matched your result! You are just Brilliant and Genius naturally.

So adjusting your integral (I thought it would require multiple integration)

If you want to find the volume of a \(\displaystyle 3\text{D}\) object in pure multiple integration, you must know in advance the formula that describes that object. In other words, you need to know how that solid is described by \(\displaystyle z\).

For example, the sphere is described by \(\displaystyle x^2 + y^2 + z^2 = r^2\). Then, its volume in pure multiple integration is:

\(\displaystyle V = \int_{-r}^{r}\int_{-\sqrt{r^2 - x^2}}^{\sqrt{r^2 - x^2}}\int_{-\sqrt{r^2 - x^2 - y^2}}^{\sqrt{r^2 - x^2 - y^2}} dz \ dy \ dx\)

But solving this triple integration in the Cartesian coordinate is mentally tiring. Therefore, we should think of parametrization and that leads us to the Spherical coordinate. The same volume can be described in the Spherical coordinate as:

\(\displaystyle V = \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{r}\rho^2\sin\phi \ d\rho \ d\phi \ d\theta\)

This multiple integration is still pure but now it is a piece of cake and can be solved in seconds. By PURE I mean that the \(\displaystyle 3\text{D}\) object was perfectly described by \(\displaystyle z\).

Some people like to make shortcuts when they find the volume of an object, so they take a slice of that solid, and do shorter calculations like we did for the prism. Sometimes we have to do this because there is no formula to describe the object or the formula is too complicated to be used.

I guess Calculus may not have been (absolutely) necessary

You are very right. Calculus was not necessary to solve this problem, but don't forget the OP requested this method! I myself would never be able to solve by any other methods!

lol

And this is the beauty of mathematics, one single problem can be solved with many different approaches and thinking. Some of these methods are very surprising by their simplicity. But some people know only how to use the harder methods!

after all but it was interesting to see your approach and thank you for taking the trouble to show in detail how you arrived at your results.

You are very welcome. And I also thank you a lot for spending much of your time to show me these surprising and fancy results.

I expect @dw1803 now has (all?) the information s/he was looking for so it's unlikely we'll ever see or hear from him/her again (not even to offer any thanks for the efforts expended on his/her behalf ).

Even if he/she has a full access to the forum, he/she will probably not come back again. Like everyone else did!

(Of course, time spent in recce is never a waste of time, as my old CO used to constantly remind us, and this has proved (IMNSHO, lol) an interesting little exercise with some useful results! )

I also say it is never a waste of time because I myself learnt a lot from this prism today. This was the first time to solve this object and I hope that my result was correct!

lol

View attachment 39478​

Hope that's of interest.

It is very beautiful. I still don't know how to do this. But Desmos is a great technology. I am learning!

 

[Agent Smith]

I'd create a scale model of the yard. Then I'd need a beaker of water.

Troglodyte solution.

So if original "cube" had side lengths 3 and I used as 1:3 scale, I'd find the beaker measures 1 of volume for my model. Now I scale it up by [imath]3^3 = 27[/imath] to get the correct answer.