F=⟨xyez,0,yzex⟩
div(F)=∇⋅F=∂x∂xyez+∂y∂0+∂z∂yzex=yezyex
curl(F)=∇×F=i∂x∂xyezj∂y∂0k∂z∂yzex=⟨zey,xyex,−xez⟩
fxfyfz=yzsinxy→f=−zcosxy+g(y,z)=xzsinxy→f=−zcosxy+h(x,z)=−cos(xy)→f=−zcosxy+k(x,y)
Problem 9
SA: z=x3+y2, between y=5x and y=x2
SA=∫∫D1+zx2+zy2