Question

The value of the triple integral \(\iiint (xy^2 + yz^3) \, dx \, dy \, dz\) over the region given by \(-1 \leq x \leq 1\), \(3 \leq y \leq 4\), \(0 \leq z \leq 2\), is .

Hint:

When solving triple integrals:
1. Simplify each term and evaluate step-by-step in the order of integration.
2. Check the symmetry of the integrand to simplify calculations.
3. Ensure correct limits of integration for all variables.

Answer 1

Step 1: Define the triple integral.
The integral is given as: \[ \iiint (xy^2 + yz^3) \, dx \, dy \, dz. \] The limits of integration are: \[ x: -1 \leq x \leq 1, \quad y: 3 \leq y \leq 4, \quad z: 0 \leq z \leq 2. \] The integral can be expressed as: \[ \int_{z=0}^{2} \int_{y=3}^{4} \int_{x=-1}^{1} (xy^2 + yz^3) \, dx \, dy \, dz. \] Step 2: Evaluate the integral over \( x \).
The inner integral with respect to \( x \) is: \[ \int_{x=-1}^{1} (xy^2 + yz^3) \, dx = \int_{x=-1}^{1} xy^2 \, dx + \int_{x=-1}^{1} yz^3 \, dx. \] 1. For \( \int_{x=-1}^{1} xy^2 \, dx \): \[ \int_{x=-1}^{1} xy^2 \, dx = y^2 \int_{x=-1}^{1} x \, dx = y^2 \left[ \frac{x^2}{2} \right]_{x=-1}^{1} = y^2 \left( \frac{1^2}{2} - \frac{(-1)^2}{2} \right) = y^2 (0) = 0. \] 2. For \( \int_{x=-1}^{1} yz^3 \, dx \): \[ \int_{x=-1}^{1} yz^3 \, dx = yz^3 \int_{x=-1}^{1} 1 \, dx = yz^3 [x]_{x=-1}^{1} = yz^3 (1 - (-1)) = 2yz^3. \] Thus, the integral over \( x \) is: \[ \int_{x=-1}^{1} (xy^2 + yz^3) \, dx = 0 + 2yz^3 = 2yz^3. \] Step 3: Evaluate the integral over \( y \).
The next integral is: \[ \int_{y=3}^{4} 2yz^3 \, dy = 2z^3 \int_{y=3}^{4} y \, dy = 2z^3 \left[ \frac{y^2}{2} \right]_{y=3}^{4}. \] \[ \int_{y=3}^{4} y \, dy = 2z^3 \left[ \frac{4^2}{2} - \frac{3^2}{2} \right] = 2z^3 \left[ \frac{16}{2} - \frac{9}{2} \right] = 2z^3 \left[ \frac{7}{2} \right] = 7z^3. \] Step 4: Evaluate the integral over \( z \).
The final integral is: \[ \int_{z=0}^{2} 7z^3 \, dz = 7 \int_{z=0}^{2} z^3 \, dz = 7 \left[ \frac{z^4}{4} \right]_{z=0}^{2}. \] \[ \int_{z=0}^{2} z^3 \, dz = 7 \left[ \frac{2^4}{4} - \frac{0^4}{4} \right] = 7 \left[ \frac{16}{4} \right] = 7 (4) = 28. \] Thus, the value of the triple integral is: \[ \iiint (xy^2 + yz^3) \, dx \, dy \, dz = 28. \] Conclusion: The value of the integral is \( 28 \).