Question

高等数学三重积分

Answer 1

λ=½时，E为CC1中点，连接AC交BD于点O，连接OE，则O为AC中点，OE是ΔCAC1的中位线。
λ=1/4时，CE=½，在正方形ABCD中，易知AC=√2，连接AC交BD于点O，连接OE，则OC=√2/2
一次洛必达法则,再使用导数的定义
lim(h→0) [f(x+2h)－2f(x+h)＋f(x)]/h^2
＝lim(h→0) [2f'(x+2h)－2f'(x+h)]/(2h)
＝lim(h→0) [f'(x+2h)－f'(x+h)]/h
＝lim(h→0) ｛2×[f'(x+2h)－f'(x)]/(2h)－[f'(x+h)－f'(x)]/h｝
＝2×lim(h→0)[f'(x+2h)－f'(x)]/(2h)－lim(h→0)[f'(x+h)－f'(x)]/h
＝2×f''(x)－f''(x)
＝f''(x)

Answer 2

2. M = ∫∫∫<Ω>(x+y)dv
= ∫<0,π/2>dt∫<0, 1>(rcost+rsint)rdr∫<r^2, 1>dz
= ∫<0,π/2>dt∫<0, 1>(cost+sint)r^2(1-r^2)dr
= ∫<0,π/2>(cost+sint)dt∫<0, 1>(r^2-r^4)dr
= (2/15)[sint-cost]<0,π/2> = 4/15

Answer 3

2. M = ∫∫∫(x+y)dv = ∫2>dt∫(rcost+rsint)rdr∫dz = ∫2>dt∫(cost+sint)r^2(1-r^2)dr = ∫2>(cost+sint)dt∫(r^2-r^4)dr = (2/15)[sint-cost]2> = 4/15