第2个回答 2018-08-20
10. 联立解 y = ax^2, y = 1-x^2, 得交点 A(1/√(a+1), a/(a+1))
直线 OA 方程 y = ax/√(a+1),
V = π∫<0, 1/√(a+1)>[(1-x^2)^2 - (ax)^2/(a+1)]dx
= π∫<0, 1/√(a+1)>{1-[2+a^2/(a+1)]x^2+x^4}dx
= π[x-(1/3){2+a^2/(a+1)}x^3+x^5/5]<0, 1/√(a+1)>
= π[1/√(a+1)-(1/3){2+a^2/(a+1)}/(a+1)^(3/2)+(1/5)/(a+1)^(5/2)]
= π[1/√(a+1) - (2/3)/(a+1)^(3/2) - (1/3)a^2/(a+1)^(5/2)+(1/5)/(a+1)^(5/2)]
= π[1/√(a+1) - (2/3)/(a+1)^(3/2) + [1/5-(1/3)a^2]/(a+1)^(5/2)]
dV/da = π{(-1/2)/(a+1)^(3/2) + 1/(a+1)^(5/2) + [(1/6)a^2-2a/3-1/2]/(a+1)^(7/2)}
令 dV/da = 0, 得 (-1/2)(a+1)^2 + a+1 + (1/6)a^2-2a/3-1/2 = 0,
即 -3(a+1)^2 + 6a+6 + a^2-4a-3 = 0,
-2a^2 - 4a = 0, a > 0 时无解。题目应有误!本回答被提问者采纳