Evaluate ∫∫∫Bf(x,y,z)dV
for the specified function f
and B: f(x,y,z)=zx3≤x≤6,0≤y≤8,0≤z≤2
∫∫∫Bf(x,y,z)dV=
To evaluate the triple integral, we integrate the function f(x, y, z) over the region B.
∫∫∫B f(x, y, z) dV = ∫∫∫B z*x^3 dV
First, we need to determine the limits of integration for each variable:
For x: x ranges from 3 to 6.
For y: y ranges from 0 to 8.
For z: z ranges from 0 to 2.
Now, we can evaluate the triple integral using these limits:
∫∫∫B z*x^3 dV = ∫∫∫B z*x^3 dz dy dx
The innermost integral will be taken with respect to z:
∫∫∫B z*x^3 dz dy dx = ∫∫ (0 to 2) [∫ (0 to 8) [∫ (3 to 6) (z*x^3) dx] dy] dz
Now, we integrate with respect to x:
∫∫ (0 to 2) [∫ (0 to 8) [∫ (3 to 6) (z*x^3) dx] dy] dz
= ∫∫ (0 to 2) [∫ (0 to 8) (z*(x^4/4)) | (3 to 6) dy] dz
= ∫∫ (0 to 2) [∫ (0 to 8) (z*(81/4)) dy] dz
= ∫∫ (0 to 2) (z*(81/4)*8) dz
= ∫ (0 to 2) (648/4)*z dz
= (648/4) * [z^2/2] (0 to 2)
= (648/4) * [(2^2/2) - (0^2/2)]
= (648/4) * (2)
= 648
Therefore, ∫∫∫B f(x, y, z) dV = 648.
To evaluate the triple integral ∫∫∫Bf(x,y,z)dV, we need to compute the integral over the region B, which is defined by the inequalities:
3≤x≤6
0≤y≤8
0≤z≤2
The function inside the integral is given by f(x,y,z) = zx^3.
We can calculate the integral by using the order of integration dz dy dx.
We need to evaluate the integral step by step.
1. Start with the innermost integral with respect to z:
∫(0 to 2) zx^3 dz
Integrating with respect to z, we get:
[z * (x^3) * (z/4)] evaluated from 0 to 2
Substituting the limits and simplifying, we get:
2 * (x^3) * (2/4) - 0 * (x^3) * (0/4)
= (x^3)
2. Now, we move on to the next integral with respect to y:
∫(0 to 8) (x^3) dy
Integrating with respect to y, we get:
(x^3) * y evaluated from 0 to 8
= 8 * (x^3)
3. Finally, we evaluate the outermost integral with respect to x:
∫(3 to 6) 8(x^3) dx
Integrating with respect to x, we get:
(8/4) * (x^4) evaluated from 3 to 6
= (8/4) * [(6^4) - (3^4)]
= (8/4) * (1296 - 81)
= (8/4) * 1215
= 8 * 1215
= 9720
Therefore, the value of the triple integral ∫∫∫Bf(x,y,z)dV is 9720.