F(x,y,z)=yzi+(xz+2yz)j+(xy+y^2)k let F = G F_x dx = xyz +c1(y.pdf
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F(x,y,z)=yzi+(xz+2yz)j+(xy+y^2)k let F = G\' F_x dx = xyz +c1(y,z) F_ydy = xyz + y^2 z +c2(x,z) F_z dz = xyz + zy^2 + c3(z,y) therefore function = xyz + zy^2 + constant (where constant is independent of x,y and z) Solution F(x,y,z)=yzi+(xz+2yz)j+(xy+y^2)k let F = G\' F_x dx = xyz +c1(y,z) F_ydy = xyz + y^2 z +c2(x,z) F_z dz = xyz + zy^2 + c3(z,y) therefore function = xyz + zy^2 + constant (where constant is independent of x,y and z) .
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F(x,y,z)=yzi+(xz+2yz)j+(xy+y^2)k let F = G F_x dx = xyz +c1(y.pdf
- 1. F(x,y,z)=yzi+(xz+2yz)j+(xy+y^2)k let F = G' F_x dx = xyz +c1(y,z) F_ydy = xyz + y^2 z +c2(x,z) F_z dz = xyz + zy^2 + c3(z,y) therefore function = xyz + zy^2 + constant (where constant is independent of x,y and z) Solution F(x,y,z)=yzi+(xz+2yz)j+(xy+y^2)k let F = G' F_x dx = xyz +c1(y,z) F_ydy = xyz + y^2 z +c2(x,z) F_z dz = xyz + zy^2 + c3(z,y) therefore function = xyz + zy^2 + constant (where constant is independent of x,y and z)