Trouble w/ deriving volumes: Frustum of pyramid w/ square base b, sqr top a, height h

[awesomest47]

Hi everyone,

I'm having a hard time with understanding how to use calculus to find the volume of a solid. It's usually most intuitive to understand in the context of a slope but my answers aren't correct. Does anyone know what I'm doing wrong? I appreciate the help.


Frustum of a pyramid with square base b, square top a and height h:

y= (b/2)-(a/2)/h (x) + b/2
We square this and take the integral of that:
1/3((b/2)-(a/2)/h (x) + b/2)^3 * h/(b/2)-(a/2)
Then we plug in h for x and do the top minus bottom number of integrand:
1/3((b/2)-(a/2)+(b/2)^3 * h/(b/2)-(a/2) - 1/3(b/2)^3 * h/(b/2)-(a/2)

But that's not correct

Another issue I had using the same approach to find the volume of a tetrahedron with side a and height h
y=a/2h (x)
Square and take integral:
1/3(a/2h)^3 * h/(a/2)
Plug in h for x and top minus bottom:
1/3(a/2)^3 * h/(a/2)

Which isn't correct either. What am I doing wrong? Thanks

 

[Steven G]

Hi everyone,

I'm having a hard time with understanding how to use calculus to find the volume of a solid. It's usually most intuitive to understand in the context of a slope but my answers aren't correct. Does anyone know what I'm doing wrong? I appreciate the help.


Frustum of a pyramid with square base b, square top a and height h:

y= (b/2)-(a/2)/h (x) + b/2
We square this and take the integral of that:
1/3((b/2)-(a/2)/h (x) + b/2)^3 * h/(b/2)-(a/2)
Then we plug in h for x and do the top minus bottom number of integrand:
1/3((b/2)-(a/2)+(b/2)^3 * h/(b/2)-(a/2) - 1/3(b/2)^3 * h/(b/2)-(a/2)

But that's not correct

Another issue I had using the same approach to find the volume of a tetrahedron with side a and height h
y=a/2h (x)
Square and take integral:
1/3(a/2h)^3 * h/(a/2)
Plug in h for x and top minus bottom:
1/3(a/2)^3 * h/(a/2)

Which isn't correct either. What am I doing wrong? Thanks

Do you really mean y= (b/2)-(a/2)/h (x) + b/2 or maybe y= [(b/2)-(a/2)]/h (x) + b/2 or maybe y= [(b/2)-(a/2)]/h (x) + b/2 or [(b/2)-(a/2)]/[h (x)] +b/2 or maybe even h (x) is a function of x or something else????

 

[Dr.Peterson]

Hi everyone,

I'm having a hard time with understanding how to use calculus to find the volume of a solid. It's usually most intuitive to understand in the context of a slope but my answers aren't correct. Does anyone know what I'm doing wrong? I appreciate the help.

Frustum of a pyramid with square base b, square top a and height h:

y= (b/2)-(a/2)/h (x) + b/2
We square this and take the integral of that:
1/3((b/2)-(a/2)/h (x) + b/2)^3 * h/(b/2)-(a/2)
Then we plug in h for x and do the top minus bottom number of integrand:
1/3((b/2)-(a/2)+(b/2)^3 * h/(b/2)-(a/2) - 1/3(b/2)^3 * h/(b/2)-(a/2)

But that's not correct

It would be easier to follow if you told us how you got that equation for y, and if you stated the entire integral (from 0 to h?). The way I did it (a line through points (0,a/2) and (h,b/2)) I got y= [(b/2)-(a/2)]/h*(x) + a/2 ; if you swapped a and b, there would be a sign error.

Other parts of your work have other missing parentheses, though I think you meant the right thing.

I would do more simplification in the midst of the work, and there is a lot to be done at the end.

How do you know, without having simplified, that your answer is wrong? My guess is that when you finish, you will find either that it is correct, or that there is a small sign error.

 

[awesomest47]

It would be easier to follow if you told us how you got that equation for y, and if you stated the entire integral (from 0 to h?). The way I did it (a line through points (0,a/2) and (h,b/2)) I got y= [(b/2)-(a/2)]/h*(x) + a/2 ; if you swapped a and b, there would be a sign error.

Other parts of your work have other missing parentheses, though I think you meant the right thing.

I would do more simplification in the midst of the work, and there is a lot to be done at the end.

How do you know, without having simplified, that your answer is wrong? My guess is that when you finish, you will find either that it is correct, or that there is a small sign error.

You're right- the b/2 was supposed to be a/2. I've been plugging in numbers for a, b and h to my answer but it doesn't match the answer that the same numbers give when entered into the known volume formula.

1/3*[(b/2]^3 * [h]/[(b/2)-(a/2)] - 1/3(a/2)^3* [h]/[(b/2)-(a/2)] is the corrected equation I have. I'm sorry with the formatting, I'm having trouble entering quotients. The integral that I took is from 0 to h and I placed the frustum on the x-axis so that the top side is closest to (0,0)

 

[awesomest47]

Ok, I tried doing some simplifying to see if it would help my volume of a tetrahedron equation:

h=a* sqrt(6)/3

1/3*(a)^3*sqrt(6)/3=
sqrt(6)*(a)^3/ 9

 

[Dr.Peterson]

You're right- the b/2 was supposed to be a/2. I've been plugging in numbers for a, b and h to my answer but it doesn't match the answer that the same numbers give when entered into the known volume formula.

1/3*[(b/2]^3 * [h]/[(b/2)-(a/2)] - 1/3(a/2)^3* [h]/[(b/2)-(a/2)] is the corrected equation I have. I'm sorry with the formatting, I'm having trouble entering quotients. The integral that I took is from 0 to h and I placed the frustum on the x-axis so that the top side is closest to (0,0)

I think this formula is correct, though it can be simplified a lot more! Maybe show us the wrong numbers you get, and the known formula you are comparing to. But simplifying should be your main task.

 

[HallsofIvy]

What do you mean by "square top b square bottom a"? That the top is a square with side length b and the bottom is a square with side length a? If so, each cross section at height z is a square with side length a linear function of z: s= pz+ q. When z= 0, the bottom, s= q= a. When z= h, the top, s= ph+ q= ph+ a= b so p= (b- a)/h. At height z, the side length is (b-a)(z/h)+ a. Imagine slicing the pyramid into "layers", each with thickness \(\displaystyle \Delta z\). A layer at height z would have area \(\displaystyle ((b\,- \,a)(z/h)\,+\, a)^2\) and volume \(\displaystyle ((b\,- a)(z/h)\,+\, a)^2\,\Delta z\). The entire volume would be the sum of those. In the limit, as \(\displaystyle \Delta z\) becomes infinitesimal, that gives the integral \(\displaystyle \int_0^h\, ((b\,- \,a)(z/h)\,+ \,a)^2\,dz\).

\(\displaystyle \displaystyle \int_0^h\, ((b\,-\, a)(z/h)\,+ \,a)^2\,dz\,= \,\int_0^h\, \left(\frac{(b\,- \,a)^2}{h^2}z^2\,+ \,\frac{2a(b\,- \,a)}{h}z\,+ \,a^2\right)\,dz\)

. . . .\(\displaystyle \displaystyle=\, \left[\frac{(b\,- \,a)^2}{3h^2}z^3\,+\, \frac{a(b\,-\,a)}{h}z^2\,+\, a^2z\right]_0^h\,= \,\frac{(b\,-\,a)^2}{3}h\,+\, a(b\,-\, a)h\,+ \,a^2h\)

. . . .\(\displaystyle \displaystyle=\, \left(\frac{b^2\,- \,2ab\,+\, a^2\,+ \,3ab\,- \,3a^2\,+\, 3a^2}{3}\right)h\,=\, \left(\frac{b^2\,+ \,ab\,+\, a^2}{3}\right)h\).

 

[awesomest47]

I think this formula is correct, though it can be simplified a lot more! Maybe show us the wrong numbers you get, and the known formula you are comparing to. But simplifying should be your main task.

So I was using hypothetical values of b=2, a=3 and h=1 and was getting:

1/3[*1]^3 *1/(1-1.5) - 1/3[(1.5)]^3 * 1/(1-1.5)

-.66 - (-2.25)= 1.59

Which is incorrect compared to the volume formula using:
1/3h (B1+B2+sqrt (B1B2)

2+3+sqrt(6)= 7.449
7.449*1*.33=2.48

Which led me to think I had done something wrong with my formula

 

[awesomest47]

What do you mean by "square top b square bottom a"? That the top is a square with side length b and the bottom is a square with side length a? If so, each cross section at height z is a square with side length a linear function of z: s= pz+ q. When z= 0, the bottom, s= q= a. When z= h, the top, s= ph+ q= ph+ a= b so p= (b- a)/h. At height z, the side length is (b-a)(z/h)+ a. Imagine slicing the pyramid into "layers", each with thickness \(\displaystyle \Delta z\). A layer at height z would have area \(\displaystyle ((b\,- \,a)(z/h)\,+\, a)^2\) and volume \(\displaystyle ((b\,- a)(z/h)\,+\, a)^2\,\Delta z\). The entire volume would be the sum of those. In the limit, as \(\displaystyle \Delta z\) becomes infinitesimal, that gives the integral \(\displaystyle \int_0^h\, ((b\,- \,a)(z/h)\,+ \,a)^2\,dz\).

\(\displaystyle \displaystyle \int_0^h\, ((b\,-\, a)(z/h)\,+ \,a)^2\,dz\,= \,\int_0^h\, \left(\frac{(b\,- \,a)^2}{h^2}z^2\,+ \,\frac{2a(b\,- \,a)}{h}z\,+ \,a^2\right)\,dz\)

. . . .\(\displaystyle \displaystyle=\, \left[\frac{(b\,- \,a)^2}{3h^2}z^3\,+\, \frac{a(b\,-\,a)}{h}z^2\,+\, a^2z\right]_0^h\,= \,\frac{(b\,-\,a)^2}{3}h\,+\, a(b\,-\, a)h\,+ \,a^2h\)

. . . .\(\displaystyle \displaystyle=\, \left(\frac{b^2\,- \,2ab\,+\, a^2\,+ \,3ab\,- \,3a^2\,+\, 3a^2}{3}\right)h\,=\, \left(\frac{b^2\,+ \,ab\,+\, a^2}{3}\right)h\).

I have the same answer for the first step but I'm somewhat confused about integrating that formula. I tried using u-substitution:
u=(x)[b/2-a/2]/[h] + a/2
So dx= du ([h]/[b/2-a/2]
And then integrating u gives
1/3 (u}^3 * [h]/[b/2-a/2]
Plug in u to get 1/3((x)[(b/2-a/2]/[h]) + a/2)^3 *[h]/[b/2-a/2]

 

[Dr.Peterson]

So I was using hypothetical values of b=2, a=3 and h=1 and was getting:

1/3[*1]^3 *1/(1-1.5) - 1/3[(1.5)]^3 * 1/(1-1.5)

-.66 - (-2.25)= 1.59

Which is incorrect compared to the volume formula using:
1/3h (B1+B2+sqrt (B1B2)

2+3+sqrt(6)= 7.449
7.449*1*.33=2.48

Which led me to think I had done something wrong with my formula

Actually it is your "correct" formula that is wrong (as you are using it). Your formula is wrong too, but not far off.

Here is the formula: V = h/3 (A₁ + A₂ + sqrt(A₁A₂), where A₁ is the top area and A₂ is the bottom area.

Note that the bases are areas, not sides! So for your example, with b=2, a=3, and h=1, your B₁ = 4 and B₂ = 9, so the formula gives V = 1/3 h (B1+B2+sqrt (B1B2)) = 1/3 (1) (4 + 9 + sqrt(4*9)) = 1/3 (19) = 19/3 = 6 1/3.

Your formula, using exact fractions rather than rounding, gives -2/3 + 9/4 = -8/12 + 27/12 = 19/12 = 1 7/12.

Do you see how these are related? I missed one thing when I said your work looked right: You forgot that you were calculating only 1/4 of the volume (by squaring your y, rather than 2y, in the integrand), so you have to multiply your formula by 4 to correct it.

In fact, if you simplify your formula, as I have been urging you to do, you get V = h/12 (a^2 + ab + b^2), which is just 1/4 of the "real" formula, when you recognize that B₁ = a² and B₂ = b².

 

[awesomest47]

Actually it is your "correct" formula that is wrong (as you are using it). Your formula is wrong too, but not far off.

Here is the formula: V = h/3 (A₁ + A₂ + sqrt(A₁A₂), where A₁ is the top area and A₂ is the bottom area.

Note that the bases are areas, not sides! So for your example, with b=2, a=3, and h=1, your B₁ = 4 and B₂ = 9, so the formula gives V = 1/3 h (B1+B2+sqrt (B1B2)) = 1/3 (1) (4 + 9 + sqrt(4*9)) = 1/3 (19) = 19/3 = 6 1/3.

Your formula, using exact fractions rather than rounding, gives -2/3 + 9/4 = -8/12 + 27/12 = 19/12 = 1 7/12.

Do you see how these are related? I missed one thing when I said your work looked right: You forgot that you were calculating only 1/4 of the volume (by squaring your y, rather than 2y, in the integrand), so you have to multiply your formula by 4 to correct it.

In fact, if you simplify your formula, as I have been urging you to do, you get V = h/12 (a^2 + ab + b^2), which is just 1/4 of the "real" formula, when you recognize that B₁ = a² and B₂ = b².

Thank you so much for your help. I apologize for misinterpreting the volume formula and not simplifying earlier- it seemed very involved so I was waiting to see if I actually had derived the right formula to begin with. One last question. When applying the principle outlined above (regarding dividing by 4) to the volume of a tetrahedron equation, I'm curious why it's not working out

y= [a/h](x)
Square and take the integral
1/3([a])^3 * h/a
We don't need to divide by anything this time because I used a instead of a/2 to begin with (I'm still kinda confused why I only found 1/4 the volume)
h=sqrt(3)/2 *a
Therefore, v= sqrt(3)/6 * [a]^3

But when plugging in 2 for a, I get 2.3 instead of .9428

 

[Dr.Peterson]

Thank you so much for your help. I apologize for misinterpreting the volume formula and not simplifying earlier- it seemed very involved so I was waiting to see if I actually had derived the right formula to begin with. One last question. When applying the principle outlined above (regarding dividing by 4) to the volume of a tetrahedron equation, I'm curious why it's not working out

y= [a/h](x)
Square and take the integral
1/3([a])^3 * h/a
We don't need to divide by anything this time because I used a instead of a/2 to begin with (I'm still kinda confused why I only found 1/4 the volume)
h=sqrt(3)/2 *a
Therefore, v= sqrt(3)/6 * [a]^3

But when plugging in 2 for a, I get 2.3 instead of .9428

It's very important to state what you are doing -- both for the sake of communication, and to make sure you are doing the right thing. It's also important to define your variables.

I THINK you are finding the volume of a square pyramid here. I suppose h is the height and a is the side of the base. Then what is y? It would be the length of a side of a cross-section x units from the apex. So it is appropriate to integrate the area of this cross-section, which is the square of the side. If that's the integral you did, I don't see how you got your result, but it does simplify to the right formula.

I don't know what the last part of your work, starting with h=sqrt(3)/2 *a, refers to. Please tell me!

As for the 1/4, please go back and tell me why you used a/2 and b/2 in the first place. What was your thinking? My assumption has been that you were finding an equation for the line forming part of the cross-section, much as one does with solids of revolution. But in the latter case, the integrand is pi y^2, y being the radius. In your case, the integrand is not y^2, but (2y)^2, because the side of the cross-section is 2y.

Finally, about simplifying: there are situations where I hold off on simplifying for a specific reason, but almost always I simplify immediately, at each step of my work, if only because it reduces the chances of error. Complicated expressions are risky.

 

[awesomest47]

It's very important to state what you are doing -- both for the sake of communication, and to make sure you are doing the right thing. It's also important to define your variables.

I THINK you are finding the volume of a square pyramid here. I suppose h is the height and a is the side of the base. Then what is y? It would be the length of a side of a cross-section x units from the apex. So it is appropriate to integrate the area of this cross-section, which is the square of the side. If that's the integral you did, I don't see how you got your result, but it does simplify to the right formula.

I don't know what the last part of your work, starting with h=sqrt(3)/2 *a, refers to. Please tell me!

As for the 1/4, please go back and tell me why you used a/2 and b/2 in the first place. What was your thinking? My assumption has been that you were finding an equation for the line forming part of the cross-section, much as one does with solids of revolution. But in the latter case, the integrand is pi y^2, y being the radius. In your case, the integrand is not y^2, but (2y)^2, because the side of the cross-section is 2y.

Finally, about simplifying: there are situations where I hold off on simplifying for a specific reason, but almost always I simplify immediately, at each step of my work, if only because it reduces the chances of error. Complicated expressions are risky.

The assignment involves finding the volume of a pyramid with a triangular base. In the last problem (involving the frustum), I "lined" the solid so that the height (a line going through the center) was on the x-axis. That way, the end of the frustum wasn't the full length of the side, but instead the portion above the middle (hence a/2). I found the h=sqrt(3)/2 *a formula online for the height of a pyramid with a triangular base and the correct formula I'm using to find the volume is v=[a]^3/6(sqrt(2))

So when I plug in a=2 to the above formula, as well as the one I derived, I get different answers

 

[Dr.Peterson]

When applying the principle outlined above (regarding dividing by 4) to the volume of a tetrahedron equation, I'm curious why it's not working out

y= [a/h](x)
Square and take the integral
1/3([a])^3 * h/a
We don't need to divide by anything this time because I used a instead of a/2 to begin with (I'm still kinda confused why I only found 1/4 the volume)
h=sqrt(3)/2 *a
Therefore, v= sqrt(3)/6 * [a]^3

But when plugging in 2 for a, I get 2.3 instead of .9428

The assignment involves finding the volume of a pyramid with a triangular base.

In the last problem (involving the frustum), I "lined" the solid so that the height (a line going through the center) was on the x-axis. That way, the end of the frustum wasn't the full length of the side, but instead the portion above the middle (hence a/2).

I found the h=sqrt(3)/2 *a formula online for the height of a pyramid with a triangular base and the correct formula I'm using to find the volume is v=[a]^3/6(sqrt(2))

So when I plug in a=2 to the above formula, as well as the one I derived, I get different answers

I assume you now understand why squaring y was wrong, because like a, y is only half the side of a square?

In this triangular pyramid problem, what does y= [a/h](x) represent? and why do you say you squared it, when the cross-section is an (equilateral?) triangle, not a square? and why do you say h=sqrt(3)/2 *a, which gives the height of an equilateral triangle, not of a tetrahedron?

Please state this whole problem and the main points of your work from start to finish, so I don't have to try to figure out what you are doing. Don't just look things up; derive them for yourself (or, at least, give your references so we can check whether you have misinterpreted something).

 

[awesomest47]

I assume you now understand why squaring y was wrong, because like a, y is only half the side of a square?

In this triangular pyramid problem, what does y= [a/h](x) represent? and why do you say you squared it, when the cross-section is an (equilateral?) triangle, not a square? and why do you say h=sqrt(3)/2 *a, which gives the height of an equilateral triangle, not of a tetrahedron?

Please state this whole problem and the main points of your work from start to finish, so I don't have to try to figure out what you are doing. Don't just look things up; derive them for yourself (or, at least, give your references so we can check whether you have misinterpreted something).

y=[a/h](x) is the equation for the slope of an edge of the pyramid. I squared it because that's what the textbook has been doing with every other slope of a line in such volume formula derivations.

I honestly have no idea what the theory behind any of these derivations are. I've been following the steps of identifying the line formula, squaring this value, taking the integral of that square, and then plugging in the values for the integrand. That's what the textbook has been doing for every problem and it was working. I really appreciate your patience though.

The problem states: Find the volume of the described solid s: a pyramid with height h and base an equilateral triangle with side a (a tetrahedron)

So following the procedure that I've been going through
1) The height is a line going through the center of the pyramid. If we place the pyramid on the x-axis over this line (where the vertex is pointing towards the left end), the y-intercept is 0, the rise is a and the run is h y=a/h(x) +0
2) Now we square this y=[(a/h)(x)]^2
3) This looks frightening to integrate so we use u-substitution.
u=[(a/h)(x)]
du= [a/h] dx
du* h/a= dx
y= 1/3(u)^3 * h/a
Plug in u to get y= 1/3(a/h(x))^3 * h/a
4) We stick in h for x so
y= 1/3(a)^3 * h/a
When x is 0, y is 0
so 1/3(a)^3 * h/a- 0

I guess this is the volume of a tetrahedron??

I really can't explain any more of my reasoning beyond this, sorry- I just lack an understanding of much of the theory of calculus.

 

[Dr.Peterson]

y=[a/h](x) is the equation for the slope of an edge of the pyramid. I squared it because that's what the textbook has been doing with every other slope of a line in such volume formula derivations.

I honestly have no idea what the theory behind any of these derivations are. I've been following the steps of identifying the line formula, squaring this value, taking the integral of that square, and then plugging in the values for the integrand. That's what the textbook has been doing for every problem and it was working. I really appreciate your patience though.

The problem states: Find the volume of the described solid s: a pyramid with height h and base an equilateral triangle with side a (a tetrahedron)

So following the procedure that I've been going through
1) The height is a line going through the center of the pyramid. If we place the pyramid on the x-axis over this line (where the vertex is pointing towards the left end), the y-intercept is 0, the rise is a and the run is h y=a/h(x) +0
2) Now we square this y=[(a/h)(x)]^2
3) This looks frightening to integrate so we use u-substitution.
u=[(a/h)(x)]
du= [a/h] dx
du* h/a= dx
y= 1/3(u)^3 * h/a
Plug in u to get y= 1/3(a/h(x))^3 * h/a
4) We stick in h for x so
y= 1/3(a)^3 * h/a
When x is 0, y is 0
so 1/3(a)^3 * h/a- 0

I guess this is the volume of a tetrahedron??

I really can't explain any more of my reasoning beyond this, sorry- I just lack an understanding of much of the theory of calculus.

If I were tutoring you face to face, I would be picking up your textbook and going through some examples with you, discussing why they do what they do. Blindly copying examples without understanding is not good.

Unfortunately, I don't have time at the moment to dig deep with you, even if writing could match sitting down together. I'll just make some comments. You should also read through the earlier examples in your book, which ought to say more than the later ones about why they do what they do.

First, in this kind of problem, the integrand is the area of a cross-section. It will be proportional to the square of some linear measure; but the area of a triangle, which is what you need here, is not the square of the side. You need to work out what that area is. Find its height as a function of the side, and use bh/2.

Second, your y is actually the length of a side of the triangle (not some distance from the centerline of the pyramid, and not the side of a square). So what you need to do is to find a formula for the area of an equilateral triangle in terms of its side, and put your y in for that side. That will give you the integrand.

Third, the integration will not be scary if you simplify first. If it really were [(a/h)(x)]^2, you could rewrite it as a^2/h^2 x^2, where a^2/h^2 is just a constant, so it can be pulled outside of the integral. Then you are just integrating x^2, which is easy. No u substitution is really needed, though it's not wrong.

 

[awesomest47]

If I were tutoring you face to face, I would be picking up your textbook and going through some examples with you, discussing why they do what they do. Blindly copying examples without understanding is not good.

Unfortunately, I don't have time at the moment to dig deep with you, even if writing could match sitting down together. I'll just make some comments. You should also read through the earlier examples in your book, which ought to say more than the later ones about why they do what they do.

First, in this kind of problem, the integrand is the area of a cross-section. It will be proportional to the square of some linear measure; but the area of a triangle, which is what you need here, is not the square of the side. You need to work out what that area is. Find its height as a function of the side, and use bh/2.

Second, your y is actually the length of a side of the triangle (not some distance from the centerline of the pyramid, and not the side of a square). So what you need to do is to find a formula for the area of an equilateral triangle in terms of its side, and put your y in for that side. That will give you the integrand.

Third, the integration will not be scary if you simplify first. If it really were [(a/h)(x)]^2, you could rewrite it as a^2/h^2 x^2, where a^2/h^2 is just a constant, so it can be pulled outside of the integral. Then you are just integrating x^2, which is easy. No u substitution is really needed, though it's not wrong.

I figured it out using the known side lengths of a 30-60-90 triangle. Thank you for your help and patience!

 

[Dr.Peterson]

I figured it out using the known side lengths of a 30-60-90 triangle. Thank you for your help and patience!

Excellent. That's the one key idea I left unstated, and you found it.