Answer :
the integral is:∬s(x y z)ds = (4/3 - 24/5 + 2/7 - 2/9) = -6/45
∬s(x y z)ds = ∬r(u,v)(2u v)(u − 2v)(u 3v)dudv
= ∫0^1∫0^2 (4uv^2 - 8uv^3 + u^2v^4 -2u^2v^5)dudv
= (4/3 - 24/5 + 2/7 - 2/9) = -6/45
We can evaluate the integral by first rewriting it in terms of parametric equations.
The surface s is defined parametrically by r(u,v) = (2u v)i (u − 2v)j (u 3v)k for 0 ≤ u ≤ 1, and 0 ≤ v ≤ 2.
Therefore, the integral can be written as ∬s(x y z)ds = ∬r(u,v)(2u v)(u − 2v)(u 3v)dudv.
Next, we can evaluate the integral by using double integrals.
We can evaluate the integral by first evaluating the inner integral with respect to v, and then evaluating the outer integral with respect to u.
The inner integral with respect to v is:
∫0^2 (4uv^2 - 8uv^3 + u^2v^4 -2u^2v^5)dv
= (4/3 - 24/5 + 2/7 - 2/9)
The outer integral with respect to u is:
∫0^1 (4/3 - 24/5 + 2/7 - 2/9)du = (4/3 - 24/5 + 2/7 - 2/9)
Therefore, the integral is:
∬s(x y z)ds = (4/3 - 24/5 + 2/7 - 2/9) = -6/45
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