Triple Integrals Over General Regions

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SUMMARY

The discussion centers on evaluating the triple integral of the function xy over the solid tetrahedron defined by the vertices (0,0,0), (1,0,0), (0,2,0), and (0,0,3). The region E lies below the plane described by the equation 2z + 6x + 3y = 6. To derive this equation, participants suggest defining vectors between points A(1,0,0), B(0,2,0), and C(0,0,3), and then using the cross product to find the normal vector to the plane. The final equation of the plane is confirmed as 6x + 3y + 2z = 6.

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Knowledge of vector operations, specifically cross products
  • Familiarity with the equation of a plane in three-dimensional space
  • Basic skills in evaluating integrals over defined regions
NEXT STEPS
  • Study the method of evaluating triple integrals over various geometric shapes
  • Learn about the derivation of plane equations from points in three-dimensional space
  • Explore vector calculus, particularly the application of cross products
  • Investigate the use of software tools like Wolfram Alpha for visualizing integrals and planes
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, vector calculus, and geometric interpretations of integrals. This discussion is beneficial for anyone looking to solidify their understanding of triple integrals and plane equations.

wubie
Hello,

First I will post my question:

Evaluate the triple integral:

Triple integral sub E of xy dV, where E is the solid tetrahedron with vertices (0,0,0) , (1,0,0) , (0,2,0) , (0,0,3)

It has been quite a while since my last calculus course so I don't remember everything. Now here is MY question: How do I find the equation of the plane in which the region E lies below?

I know from the solution manual that the E is the region that lies below the plane

2z + 6x + 3y = 6

How do I find that out?

I found three separate equations for each plane - xy, yz, xz.

6 = 2z + 6x, 6 = 2z + 3x, 6 = 3y + 6x.

And I can see the relationship between all four planes. How do I come up with the final equation of the plane

2z + 6x + 3y = 6


I should know this. I just can't remember.

Any help is appreciated. Thankyou.
 
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Define two vectors between points A(1,0,0), B(0,2,0), and C(0,0,3). It doesn't matter which pair of points you use to make the vectors, as long as the vectors are different.

The cross product of these two vectors will be normal to the plane. Then the equation,
[tex]n_x(x-x_0) + n_y(y-y_0) + n_z(z-z_0)=0[/tex]
will describe the plane.

[tex]\boldsymbol{AB}=<1,-2,0>[/tex]
[tex]\boldsymbol{AC}=<1,0,-3>[/tex]
[tex]\boldsymbol{AB}\times\boldsymbol{AC}=<6,3,2>=\boldsymbol{n}[/tex]
[tex]6(x-1)+3y+2z=0[/tex]
[tex]6x+3y+2z=6[/tex]

cookiemonster
 
Thanks for the help cookie monster.
 

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