Volume of the frustum of a pyramid

[akbar786]

Homework Statement

Find the volume of the frustum of a pyramid with square base of side b,
square top of side a, and height h. What happens if a = b ? If a = 0 ?

Homework Equations

None

The Attempt at a Solution

I know that i have to integrate from 0 to H. I make a generic square with sides X. The area of the square would be x*x or X^2 and the height would be dx. I have to integrate that, but how can i find what exactly X is? Since the base is a larger square, i know the x should be decreasing as it moves from base b towards base a. My problem is that i cannot figure out how to get the value of x since there are no numbers given for the problem.

Also, can someone confirm this if its right or not?
if a=b then the bottom square and the top square become the same length so the shape would become a rectangular box right?
If a = 0 then the shape would just become a regular pyramid rather than a frustum of a pyramid?

 

[Mark44]

Homework Statement

Find the volume of the frustum of a pyramid with square base of side b,
square top of side a, and height h. What happens if a = b ? If a = 0 ?

Homework Equations

None

The Attempt at a Solution

I know that i have to integrate from 0 to H. I make a generic square with sides X. The area of the square would be x*x or X^2 and the height would be dx.

No, the height or thickness of your volume element would be dy.

If you haven't already done so, draw a picture of the vertical cross-section of your frustum of a pyramid, with the base of the pyramid along the x-axis. The endpoints of the base should be at (-b/2, 0) and (b/2, 0). The endpoints of the top of the frustum should be at (-a/2, H) and (a/2, H). You will need equations for the slanted sides.

I have to integrate that, but how can i find what exactly X is? Since the base is a larger square, i know the x should be decreasing as it moves from base b towards base a. My problem is that i cannot figure out how to get the value of x since there are no numbers given for the problem.

Also, can someone confirm this if its right or not?
if a=b then the bottom square and the top square become the same length so the shape would become a rectangular box right?
If a = 0 then the shape would just become a regular pyramid rather than a frustum of a pyramid?

 

[akbar786]

once i find the equations of the slanted sides what exactly am i supposed to do?

 

[Mark44]

Then you can get an equation for x in terms of y. You'll be integrating with respect to y, so you need x in terms of y.

 

[akbar786]

maybe I am not understanding your answer correctly. This is what i have for y so far. y = 2(b/2 +(a/2-b/2)*y/h). I put the two outside so whatever the x value comes out to be it will be multiplied by 2 that way i get the whole length and not just half of it. If h = 0 then y = 0 , thus we are left with 2b/2 or just b, if y = h then its just 2(b/2+(a/2-b/2). Which comes out to be b/2 + a-b/2 which results in 2a/2 or a. Since i have that in terms of y could i integrate that from 0 to h? The integral would be o to h of (2(b/2+(1/2-b/2)) ^2. Since its a square in order to get the area i just squared the side. Is it right so far?

 

[Mark44]

maybe I am not understanding your answer correctly. This is what i have for y so far. y = 2(b/2 +(a/2-b/2)*y/h).

This equation makes no sense at all, since it does not include x. Any line that is neither vertical nor horizontal will have an equation that includes two variables, usually x and y.

Since you are attempting to work an calculus problem, you should know how to find the equation of a line, given two points on the line. In this case, the two points are (b/2, 0) and (a/2, H).

When you get the equation of the line, it will probably be y = <an expression involving x>. Solve that equation so that you get x as a function of (in terms of) y.

I put the two outside so whatever the x value comes out to be it will be multiplied by 2 that way i get the whole length and not just half of it. If h = 0 then y = 0 , thus we are left with 2b/2 or just b, if y = h then its just 2(b/2+(a/2-b/2). Which comes out to be b/2 + a-b/2 which results in 2a/2 or a. Since i have that in terms of y could i integrate that from 0 to h? The integral would be o to h of (2(b/2+(1/2-b/2)) ^2. Since its a square in order to get the area i just squared the side. Is it right so far?