Answer to Question #198649 in Calculus for anuj

Question #198649

Let 𝐹 = 2π‘₯𝑧𝑖 βˆ’ π‘₯𝑗 + 𝑦 2π‘˜. Evaluate 𝑉 𝐹𝑑𝑉 is the region bounded by the surfaces π‘₯ = 0, 𝑦 = 0, 𝑧 = π‘₯ 2 , 𝑧 = 4.


1
Expert's answer
2022-08-04T11:07:54-0400

"F=2xzi-xj+y^2k"

The region bounded by the surfaces: "x=0,y=0,z=x^2+y^2,z=4"

(ModificationΒ "z=x^2+y^2"Β was made, because the region is bounded in y-direction only from one side: y=0.)


"divF=\\frac{\\partial}{\\partial x}(2xz)+\\frac{\\partial}{\\partial y}(-x)+\\frac{\\partial}{\\partial z}(y^2)=2z"


Let the region bounded by the surfaces: "x=0,y=0,z=x^2+y^2,z=4"


"\\iiint divFdV=2\\int^2_0dx\\int^2_0dy\\int^4_{x^2+y^2}zdz=\\int^2_0dx\\int^2_0(16-(x^2+y^2)^2)dy="


"=\\int^2_0(16y-x^4y-2x^2y^3\/3-y^5\/5)|^2_0dx=\\int^2_0(32-2x^4-16x^2\/3-32\/5)dx="


"=(32x-2x^5\/5-16x^3\/9-32x\/5)|^2_0=64-64\/5-128\/9-64\/5="


"=1088\/45"



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