Question

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 help with finding the boundries using the graph and then solving it.

Answer 1

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 = $\left[\begin{array}{ccc}x&y&z\\0&2&-3\\1&0&-3\end{array}\right]$

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,

$0 \leq Z \leq 3-3x- \frac{3y}{2}$

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


$0 \leq y \leq -2x+2$

-------------------------------------

With  ,


$0 \leq x \leq 1$

----------------------------------------


$\\ \int\limits^1_0 {} \, \int\limits^ \frac{-2x+2}{} _0 {} \, \int\limits^ \frac{3-3x- \frac{3y}{2} }{} _0 {(xy)} \, dzdydx \\ \\ =\int\limits^1_0 {} \, \int\limits^ \frac{-2x+2}{} _0 {} \,(3xy -3x^2y - \frac{3xy^2}{2} )dydx \\ \\ =\int\limits^1_0 {} \, ( \frac{3xy^2}{2} - \frac{3x^2y^2}{2} - \frac{3xy^3}{6} )|0,(-2x+2)dx \\ \\ = \int\limits^1_0 {(\frac{3x(-2x+2)^2}{2} - \frac{3x^2(-2x+2)^2}{2} - \frac{3x(-2x+2)^3}{6} )} \, dx$

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

$\\ = \int\limits^1_0 {(\frac{3x(-2x+2)^2}{2} - \frac{3x^2(-2x+2)^2}{2} - \frac{3x(-2x+2)^3}{6} )} \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2[ 1- x- \frac{1}{3} (-2x+2)] } \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2[ 1- x+ \frac{2x}{3} - \frac{2}{3} ] } \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2[ \frac{1}{3} - \frac{x}{3}] } \, dx \\ \\ = \int\limits^1_0 { \frac{3x}{2}(-2x+2)^2( \frac{1-x}{3} ) } \, dx$

$\\ = \int\limits^1_0 { \frac{x}{2}(-2x+2)^2(1-x) } \, dx \\ \\ = \int\limits^1_0 { \frac{x}{2}(4x^2-8x+4)(1-x) } \, dx \\ \\ = \int\limits^1_0 {(2x^3-4x^2+2x) (1-x) } \, dx \\ \\ = \int\limits^1_0 {(-2x^4+4x^3-2x^2+2x^3-4x^2+2x)} \, dx \\ \\ = \int\limits^1_0 {(-2x^4+6x^3-6x^2+2x)} \, dx \\ \\ = -\frac{2x^5}{5} + \frac{6x^4}{4} - \frac{6x^3}{3} + \frac{2x^2}{2} |(0,1) \\ \\ = -\frac{2}{5} + \frac{6}{4} - \frac{6}{3} + \frac{2}{2}$

$\\ =-\frac{2}{5} + \frac{3}{2} - 2 + \frac{2}{2} \\ \\ = -\frac{2}{5} -2+ \frac{3+2}{2} \\ \\ = -\frac{2}{5} -2 + 5/2 \\ \\ = \frac{1}{10} u.v$