Let πΉ = 2π₯π§π β π₯π + π¦2π. Evaluate π πΉππ is the region bounded by the surfaces π₯ = 0, π¦ = 0, π§ = π₯2 , π§ = 4.
Here,
"F= 2xzi-xj+y^2k"
The region V is covered
(a) by keeping x and y fixed and integrating from "z=x^2 \\ to\\ z=4" (base to top of column PQ),
(b) then by keeping x fixed and integrating from y=0 to y=6 (R to S in the slab)
(c) finally integrating from x=0 to x=2 (where"z=x^2" meets z=4)
Then the required integral is :
"\\int\\int\\int_V\\ Fdv = \\int_{x=0}^2\\int_{y=0}^6\\int_{x^2}^4(2xzi-xj+y^2k)dzdydx"
"=i\\int_0^2\\int_0^6\\int_{x^2}^4(2xz)dzdydx-j\\int_0^2\\int_0^6\\int_{x^2}^4(x)dzdydx+k\\int_0^2\\int_0^6\\int_{x^2}^4(y^2)dzdydx\\\\\\ \\\\=128i-24j+384k"
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