Chapter 13: Q. 25 (page 1055)

Evaluate the iterated integral :
$\int_{0}^{1} \int_{2}^{4} \int_{- 1}^{5} (x + y z^{2}) d x d y d z$

Short Answer

Expert verified

Step by step solution

Step 1. Given information.

Given integral is :

$\int_{0}^{1} \int_{2}^{4} \int_{- 1}^{5} (x + y z^{2}) d x d y d z$

We have to evaluate it.

Step 2. Evaluate.

$\begin{aligned} = \int_{0}^{1} \int_{2}^{4} \int_{- 1}^{5} (x + y z^{2} d x) d y d z l \\ = \int_{0}^{1} \int_{2}^{4} (\int_{- 1}^{5} x d x + \int_{- 1}^{5} y z^{2} d x) d y d z \\ = \int_{0}^{1} \int_{2}^{4} ({(\frac{x^{2}}{2})}_{- 1}^{5} + y z^{2} (x)_{- 1}^{5}) d y d z \\ = \int_{0}^{1} \int_{2}^{4} ((\frac{5^{2}}{2} - \frac{(- 1)^{2}}{2}) + y z^{2} (5 - (- 1)) d y d z \\ = \int_{0}^{1} \int_{2}^{4} (12 + 6 y z^{2}) d y d z \end{aligned}$

Step 3. Integrate with respect to y.

Integrate with respect to y

$\begin{aligned} = \int_{0}^{1} (\int_{2}^{4} 12 d y + 6 z^{2} \int_{2}^{4} y d \cdot y) d z \\ = \int_{0}^{1} (12 (y)_{2}^{4} + 6 z^{2} {(\frac{y^{2}}{2})}_{2}^{4}) d z \\ = \int_{0}^{1} (12 (4 - 2) + 6 z^{2} (\frac{4^{2}}{2} - \frac{2^{2}}{2})) d z \\ = \int_{0}^{1} (24 + 36 z^{2}) d z \end{aligned}$

Integrate with respect to z

$\begin{aligned} = 24 \int_{0}^{1} d z + 36 \int_{0}^{1} z^{2} d z Algebra of \\ = 24 (z)_{0}^{1} + 36 {(\frac{z^{3}}{3})}_{0}^{1} Integ \\ = 24 (1 - 0) + 36 (\frac{1}{3} - 0) Apply lin \\ = 24 + 12 \\ = 36 \\ ∴ \int_{0}^{1} \int_{2}^{4} \int_{- 1}^{5} (x + y z^{2}) d x d y d z = 36 \end{aligned}$

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Evaluate the iterated integral :
$\int_{0}^{1} \int_{2}^{4} \int_{- 1}^{5} (x + y z^{2}) d x d y d z$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Evaluate.

Step 3. Integrate with respect to y.

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Short Answer

Step by step solution

Step 1. Given information.

Step 2. Evaluate.

Step 3. Integrate with respect to y.

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Evaluate the iterated integral :
$\int_{0}^{1} \int_{2}^{4} \int_{- 1}^{5} (x + y z^{2}) d x d y d z$