Volume of a tetrahedron by Triple Integral

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Homework Help Overview

The problem involves finding the volume of a tetrahedron using a triple integral, specifically bounded by the coordinate planes and the plane defined by the equation 2x + 3y + 2z = 6.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss finding the intercepts of the plane on the axes and express uncertainty about determining the limits of integration for the triple integral. There is mention of drawing a picture to visualize the limits and comparing the process to finding areas in 2D.

Discussion Status

The discussion is ongoing, with participants exploring different methods for setting up the triple integral. Some guidance has been provided regarding the relationship between the 3D setup and the 2D case, but no consensus has been reached on the specific limits of integration.

Contextual Notes

Participants are working within the constraints of homework rules and are focused on understanding the geometric interpretation required for setting up the integral limits.

faiz4000
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Homework Statement


By using triple integral, find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x+3y+2z=6.

Homework Equations



Volume= ∫vdv=∫∫∫dxdydz

The Attempt at a Solution



find intercepts of the plane on the axes,
x-intercept=3
y-intercept=2
z-intercept=3then i don't know how to get limits of integration in the formula
 
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faiz4000 said:

Homework Statement


By using triple integral, find the volume of the tetrahedron bounded by the coordinate planes and the plane 2x+3y+2z=6.


Homework Equations



Volume= ∫vdv=∫∫∫dxdydz

The Attempt at a Solution



find intercepts of the plane on the axes,
x-intercept=3
y-intercept=2
z-intercept=3


then i don't know how to get limits of integration in the formula

Draw a picture of the plane in the first octant by joining those 3 points. Then use that picture for the limits.
 
Yes i get that. If I were finding the area of in 2D, i would draw lines parallel to x or y-axis and find the curves between which they lie. These would be the limits of inner integral. then i find the lowest and highest value of the outer integral and that becomes the limits for it... but in 3D i would have to draw planes parallel to say xy plane...but don't know which curves they lie between.
 

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faiz4000 said:
Yes i get that. If I were finding the area of in 2D, i would draw lines parallel to x or y-axis and find the curves between which they lie. These would be the limits of inner integral. then i find the lowest and highest value of the outer integral and that becomes the limits for it... but in 3D i would have to draw planes parallel to say xy plane...but don't know which curves they lie between.

It's the same idea in 3D. If you are going to do the inner integral in the ##z## direction first, you go from ##z## on the bottom surface (the xy plane) to ##z## on the top surface (the plane). Once you have done that you will have an xy integral and you can look at the triangle in the xy plane for the limits, just as you would do in 2D.

If you wanted to go in the x direction first for some reason, you would go from x on the back surface to x on the front surface then look in the yz plane for the dydz limits.
 
got it thanx.
 

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