Finding Limits of Integration for Double Integrals: Can You Help Me?

[Derill03]

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]

Hi Derill03!

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

For fixed x, what does y vary between?

 

[Derill03]

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]

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!

 

[Derill03]

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]

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

That'll do! …

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

Area of pyramid = 1/6 (a x b).c
​

 

[Derill03]

when i do the triple scalar product i get 6 as an answer? the integration way and vector way don't agree, any thoughts on what is wrong?

 

[tiny-tim]

when i do the triple scalar product i get 6 as an answer? the integration way and vector way don't agree, any thoughts on what is wrong?

D'oh! … Area of pyramid = 1/6 (a x b).c

 

[Derill03]

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]

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

Pyramid is easier to write

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 …

I copied-and-pasted the other formula, and forgot to change "area" to "volume"

 

[Derill03]

now it all makes sense, thank you very much you were a big help