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wubie
Feb26-04, 02:01 AM
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.

cookiemonster
Feb26-04, 02:20 AM
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,
n_x(x-x_0) + n_y(y-y_0) + n_z(z-z_0)=0
will describe the plane.

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

cookiemonster

wubie
Feb26-04, 02:52 AM
Thanks for the help cookie monster.