Triple integral - solid tetradhedon

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SUMMARY

The discussion focuses on evaluating the triple integral \(\int \int \int xy \, dV\) over the solid tetrahedron defined by vertices (0,0,0), (4,0,0), (0,1,0), and (0,0,7). The normal vector \(n\) is calculated using the cross product of vectors AB and AC, resulting in \(n = \langle 7, 28, 4 \rangle\). The equation of the plane is derived as \(7x + 28y + 4z = 28\), leading to the bounds for \(z\) as \(0 \leq z \leq 28 - 7x - 28y\), \(y\) as \(0 \leq y \leq 1 - \frac{1}{4}x\), and \(x\) as \(0 \leq x \leq 1\).

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Whatupdoc
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Evaluate the triple integral \int \int \int xy*DV where E is the solid
tetrahedon with vertices (0,0,0), (4,0,0),(0,1,0),(0,0,7)

first I'm going to find n:
AB= <-4,1,0>
AC= <-4,0,7>
AB X AC = <7,28,4> = n

so i get this equation: 7(x-4) + 28y + 4z = 0
=> 7x+28y+4z = 28

so the bounds of z should be z = 0..28-7x-28y

solving for y gives me the bounds for y = 0.. 1-1/4*x

and the x bounds x = 0..1

are my bounds correct?
 
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Whatupdoc said:
...
=> 7x+28y+4z = 28
so the bounds of z should be z = 0..28-7x-28y
...
Don't forget that z had a coefficient of 4.
 

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