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Limits of integration

  • Thread starter Derill03
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The question is this:

Consider the tetrahedron which is bounded on three sides by the coordinate planes and on fourth side by plane x+(y/2)+(z/3)=1

I think the region to integrate over should appear in R^2 as a right triangle, is this correct?

Secondly i am having much trouble finding limits of integration for a double integral, can ne one help
 

tiny-tim

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Hi Derill03! :smile:

For fixed x and y, what does z vary between?

For fixed x, what does y vary between? :wink:
 
I get a double integral integrating dydx of the function (3-3x-(3y/2)) from dy|0 to 2-2x and dx|0 to 1

so that would leave a volume of 1
 

tiny-tim

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I get a double integral integrating dydx of the function (3-3x-(3y/2)) from dy|0 to 2-2x and dx|0 to 1

so that would leave a volume of 1
Looks good! :smile:
 
On the next part it says to use vector methods to check the integration answer, can you point me in the right direction as to how do i use vector methods to calculate volume?

the only volume formula i know using vectors is triple scalar product (a dot (b cross c).
 

tiny-tim

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the only volume formula i know using vectors is triple scalar product (a dot (b cross c).
That'll do! :biggrin:

Area of triangle = 1/2 (a x b)

Area of pyramid = 1/6 (a x b).c :wink:
 
when i do the triple scalar product i get 6 as an answer? the integration way and vector way dont agree, any thoughts on what is wrong?
 

tiny-tim

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when i do the triple scalar product i get 6 as an answer? the integration way and vector way dont agree, any thoughts on what is wrong?
D'oh! :rolleyes:Area of pyramid = 1/6 (a x b).c
 
I understand that if you take 1/6 of 6 you will get 1 which does agree, but its confusing to me why an area formula for a pyramid would prove a correct volume of a tetrahedron? Can you explain a little bit why this works

Is it safe to assume that a tetrahedron and a pyramid are geometrically the same?
 

tiny-tim

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Is it safe to assume that a tetrahedron and a pyramid are geometrically the same?
Pyramid is easier to write :wink:
I understand that if you take 1/6 of 6 you will get 1 which does agree, but its confusing to me why an area formula for a pyramid would prove a correct volume of a tetrahedron? Can you explain a little bit why this works
oh :rolleyes:

I copied-and-pasted the other formula, and forgot to change "area" to "volume" :smile:
 
now it all makes sense, thank you very much you were a big help
 

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