Respuesta :

Answer:

2/15

Step-by-step explanation:

given that the triple integral = ∫∫∫ 8x^2 dv

and T is the solid tetrahedron with vertices :  (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1)

hence the equation of the plane: x + y + z = 1

T [ (X,Y,Z) : 0≤x≤1, 0≤y≤1-x, 0≤z≤1-x-y ]

attached below is the detailed solution ( we multiply our answer after evaluation with the coefficient of  8 as attached to the initial expresssion)

Ver imagen batolisis

By evaluating the triple integral where T is the solid tetrahedron with verticies (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1) the value calculated is 2/15.

Given :

The triple integral ∭ 8x^2 dV, where T is the solid tetrahedron with verticies (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1).

The following calculation can be used to evaluate the triple integral:

[tex]\rm I = \int\int\int 8x^2dV[/tex]

T[(x,y,z) : [tex]0 \leq x \leq 1[/tex] ; [tex]0 \leq y \leq 1-x[/tex] ; [tex]0 \leq z \leq 1-x-y[/tex] ]

Now put the limits in the above integral.

[tex]\rm I = \int\limits^1_0\int\limits^{1-x}_0\int\limits^{1-x-y}_0 {8x^2} \, dz \, dy \, dx[/tex]

[tex]\rm I = \int\limits^1_0\int\limits^{1-x}_0 {8x^2} (1-x-y) \, dy \, dx[/tex]

[tex]\rm I = 8 \int\limits^1_0\int\limits^{1-x}_0 {x^2-x^3-x^2y} \, dy \, dx[/tex]

[tex]\rm I = 8 \int\limits^1_0 {x^2(1-x)-x^3(1-x)-x^2\dfrac{(1-x)^2}{2}} \, dx[/tex]

[tex]\rm I = 8 \int\limits^1_0 {x^2-x^3-x^3+x^4-x^2\dfrac{(1+x^2-2x)}{2}} \, dx[/tex]

[tex]\rm I = 4\left( \int\limits^1_0 {2x^2-4x^3+2x^4-x^2-x^4+2x^3} \, dx\right)[/tex]

[tex]\rm I = 4\left( \int\limits^1_0 {x^2+x^4-2x^3} \, dx\right)[/tex]

[tex]\rm I = 4\left( {\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^4}{2}} \right)^1_0[/tex]

[tex]\rm I = 4\left( {\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{2}} \right)[/tex]

[tex]\rm I = \dfrac{2}{15}[/tex]

By evaluating the triple integral where T is the solid tetrahedron with verticies (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1) the value calculated is 2/15.

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https://brainly.com/question/24308099

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