Question

The value of $\iiint_B (x + 2y + 3z)^2 \, dx \, dy \, dz$ is, where $B$ is the box $[0, 1] \times \left[-\frac{1}{2}, 0\right] \times \left[0, \frac{1}{3}\right]$

• A. 0
• B. $\frac{1}{12}$
• C. $\frac{1}{3}$
• D. 1

Hint:

For linear combination integrands $f(a x + b y + c z)$, each integration step introduces a scaling factor of $1/a, 1/b, 1/c$ from the chain rule. Here $1 \cdot 2 \cdot 3 = 6$, and dividing during integration reduces the power integral cleanly!

Answer 1

The Correct Option is B

Step 1: Concept
This is a triple integral evaluated over a 3D rectangular box $B$ defined by: \[ 0 \leq x \leq 1, \quad -\frac{1}{2} \leq y \leq 0, \quad 0 \leq z \leq \frac{1}{3} \]

Step 2: Key Formulas and Approach
We perform successive single integrations from inside to outside using substitution or power rule integration: \[ I = \int_0^1 \int_{-1/2}^0 \int_0^{1/3} (x + 2y + 3z)^2 \, dz \, dy \, dx \]

Step 3: Step-by-step Explanation

• First Integration with respect to $z$: \[ \int_0^{1/3} (x + 2y + 3z)^2 \, dz = \left[ \frac{(x + 2y + 3z)^3}{3 \cdot 3} \right]_{z=0}^{z=1/3} \] \[ = \frac{(x + 2y + 1)^3 - (x + 2y)^3}{9} \]
• Second Integration with respect to $y$: \[ \int_{-1/2}^0 \frac{(x + 2y + 1)^3 - (x + 2y)^3}{9} \, dy = \frac{1}{9} \left[ \frac{(x + 2y + 1)^4}{4 \cdot 2} - \frac{(x + 2y)^4}{4 \cdot 2} \right]_{y=-1/2}^{y=0} \] \[ = \frac{1}{72} \left[ \left( (x + 1)^4 - x^4 \right) - \left( x^4 - (x - 1)^4 \right) \right] \] \[ = \frac{1}{72} \left[ (x + 1)^4 - 2x^4 + (x - 1)^4 \right] \]
• Third Integration with respect to $x$: \[ I = \frac{1}{72} \int_0^1 \left[ (x + 1)^4 - 2x^4 + (x - 1)^4 \right] dx \] \[ = \frac{1}{72} \left[ \frac{(x + 1)^5}{5} - \frac{2x^5}{5} + \frac{(x - 1)^5}{5} \right]_0^1 \]
• Evaluate at upper limit $x = 1$: \[ \frac{1}{5} \left( 2^5 - 2(1)^5 + 0^5 \right) = \frac{32 - 2}{5} = \frac{30}{5} = 6 \]
• Evaluate at lower limit $x = 0$: \[ \frac{1}{5} \left( 1^5 - 0 + (-1)^5 \right) = \frac{1 - 1}{5} = 0 \]
• Calculate final value: \[ I = \frac{1}{72} [6 - 0] = \frac{6}{72} = \frac{1}{12} \]

Step 4: Final Answer
The value of the triple integral over the box $B$ is $\frac{1}{12}$. Thus, Option (B) is correct.