Volume of a rectangle by cross-sections

  • Thread starter brushman
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  • #1
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Homework Statement


There is no specific problem, I'm just confused after reading the chapter.

Consider a pyramid 3 m high with a square base that is 3 m on a side. The cross section of the pyramid perpendicular to the altidude x m down from the vertex is a square x m on a side.

Now I understand the volume is,

[tex]
\int_{a}^{b} A(x) dx = \int_{0}^{3} x^2 dx = 9
[/tex]

but then I get confused. How would the volume of a rectangle with the same square base, using the same method, be any different?

edit: Thanks, it makes much more sense to me now.
 
Last edited:

Answers and Replies

  • #2
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First of all, there is no such thing as the "volume of a rectangle." If you mean a rectangular prism, then the volume will be different. A rectangular prism with the same square base and height would have a volume of 27 cubic units.

[tex] V_{rectangular \ prism} = bh. [/tex]
[tex] V_{pyramid} = \frac{bh}{3}. [/tex]

(where b = area of base)
 
  • #3
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Thanks Rasko. Indeed I meant rectangular prism.

I know that the volume of a rectangular prism is just the area of the base times height, but I'd like to know the volume by the method of slicing such as in my example. That way, I can compare the two to help my understanding.
 
  • #4
193
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In the pyramid integration, x varies from 0 to 3. However, each cross section of a rectangular prism has the same base. So you would have instead:

[tex] \int_0^3 (3)^2 dx [/tex]
 
Last edited:

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