Step by step solution to "Evaluate the integral. ∫−11∫01∫02(x+y+z)dydxdz"

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Answered question

2021-11-07

Evaluate the integral.

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

Caren

Skilled2021-11-08Added 96 answers

Step 1
To evaluate the integral:

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

Evaluating the integral:

\int_{- 1}^{1} \int_{0}^{1} \int_{0}^{2} (x + y + z) d y d x d z = \int_{- 1}^{1} \int_{0}^{1} [\int_{0}^{2} (x + y + z) d x] d y d z

= \int_{- 1}^{1} \int_{0}^{1} {[\frac{x^{2}}{2} + x y + x z]}_{0}^{2} d y d z

= \int_{- 1}^{1} \int_{0}^{1} [2 + 2 y + 2 z - 0] d y d z

= 2 \int_{- 1}^{1} \int_{0}^{1} (1 + y + z) d y d z

= 2 \int_{- 1}^{1} {(y + \frac{y^{2}}{2} + y z)}_{0}^{1} d z

= 2 \int_{- 1}^{1} (1 + \frac{1}{2} + z - 0) d z

= 2 \int_{- 1}^{1} (\frac{3}{2} + z) d z

= 2 {(\frac{3}{2} z + \frac{z^{2}}{2})}_{- 1}^{1}

=(3+1-3-1)
=0
Step 2
Hence, required answer is 0.

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