Evaluate the triple integrals below where e is the solid tetrahedron with vertices (0,0,0), (1,0,0), (0,2,0) and (0,0,3). i need with finding the boundries using the graph and then solving it.

Hi, We can to calculate the vectors.

And the determinant will be the plan Z

Let  A = (0,03), B =(0,2,0) , C = (1,0,0) and D = (0,0,0)

Then,

AB = B - A

Replacing the points:

AB = (0,2,0) - (0,0,3)

AB = (0i, 2j , -3k)
----------------------------

Already the vector AC = C -A

That's is,

AC = (1,0,0) - (0,0,3)

AC = (1i, 0j, -3k)

Then,

The plan = 

Solving it, we will have:

Plan:  -6x -3y -2z + d = 0

Replacinng any point to find the value of d

Example the point A =(0,0,3)

-6(0) -3(0) -2(3) + d = 0

-6+d = 0

d = 6

Then, The us equation will stay of form following :

-6x -3y -2z +6 = 0

or

6x + 3y +2z -6 = 0

Isolating 2z:

2z = 6 -6x - 3y

Dividing both the sides od equation by 2

z = 3 - 3x - 3y/2

Then,



Now, Let's find the domain in xy

|y
|  (0,2)
|\
|  \
|    \
|       \  (1,0)    
------------------------- x

b = Cut in y

then b will be = 2

As y = ax + b

y = ax + 2

We have the point = (1,0)

Replace in the equation

0 = a(1) + 2

0 = a + 2

Isolate a

a = -2

Then us stay:

y = -2x + 2



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With  ,



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Now putting 3x/2(-2x+2)²  as commu factor

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