Question #289059

1. Evaluate (𝑥 − 3) 𝑑 where = , , − 1 ≤ ≤ 1, 0 ≤ ≤ 2, 0 ≤ ≤ 1 }.

1
Expert's answer
2022-01-20T17:13:55-0500

zyx(x3)dxdydz010211(x3)dxdydz0102x23x11dydz01026dydz01(6y02)dz0112dz=12z01=12\int_z\int_y\int_x(x-3)dxdydz\\\int_0^1\int_0^2\int_{-1}^1(x-3)dxdydz\\ \int_0^1\int_0^2x^2-3x|_{-1}^1dydz\\\int_0^1\int_0^2-6dydz\\\int_0^1(-6y|_0^2)dz\\\int_0^1-12dz=-12z|_0^1=-12\\


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