Compute the scalar surface integral ∬T(z+y)dS where T is the triangle (including its interior) with vertices (0,0,2), (0,1,0)( and (1,0,0)

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abhinavjh321 abhinavjh321 · Answer 1 · 2022-12-18T14:06:08+01:00

the scalar surface integral ∬T(z+y)dS where T is the triangle (including its interior) with vertices (0,0,2), (0,1,0)( and (1,0,0) is 1

We can compute the scalar surface integral ∬T(z+y)dS using the parametric form of the surface integral.

Let S be the surface of the triangle. We can parameterize S as follows:

S(u,v) = (u, v, 2-u-v)

where 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1-u.

Then, the scalar surface integral ∬T(z+y)dS is defined as:

∬T(z+y)dS = ∬S(u,v) (2-u-v + v) dudv

= ∫0→1∫0→1-u (2-u-v + v) dudv

= ∫0→1 [2u + (1-u)2 - (1-u)2u - u2(1-u)] du

= ∫0→1 (2u + 1 - 2u2 - u3) du

= [u2 + u - u4/4]0→1

= 1/4 + 1 - ¼

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