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Evaluate ∭ B

f(x,y,z)dV for the specified function f and B : f(x,y,z)= x
z
3≤x≤24,0≤y≤2,0≤z≤2 ∭ B
f(x,y,z)dV=

User AleXelton
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1 Answer

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Let us evaluate the triple integral ∭B f(x,y,z)dV where f(x,y,z) = xz with ranges of integration as 3≤x≤24, 0≤y≤2, and 0≤z≤2.

To solve this, we perform the integration successively over z, y, and x.

Starting with z, we integrate the function xz over the range 0 to 2 for z:

∫(from 0 to 2) xz dz. The result is xz^2 / 2 when evaluated from 0 to 2 which simplifies to 2x.

Next, we integrate the obtained result ∫(from 0 to 2) 2x dy over y. But since the function does not depend on y, the integral simplifies to 4x evaluated from 0 to 2.

Finally, we evaluate ∫(from 3 to 24) 4x dx over the range of x. This yields 2x^2 when evaluated from 3 to 24.

On subtracting the lower limit from the upper limit, we get:
= 2*(24)^2 - 2*(3)^2
= 2*576 - 2*9
= 1134

Therefore, the value of the triple integral ∭B f(x,y,z)dV over the specified function and boundaries is 1134.

User Jay Levitt
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