Question

利用二重积分求y=x+1与y^2=1-x所围成平面区域的面积

Answer 1

直线y=x+1与抛物线y^2=1-x的交点满足这两个方程: y=x+1, y^2=1-x
解得两个交点为： (0, 1) , (-3,-2) .
所以，直线y=x+1与抛物线y^2=1-x 围成的区域为 D： -2<=y<=1，y-1<=x<=1-y^2
（画图就可发现直线在D的左边界，抛物线在右边界）
所以 D的面积为 ： ∫∫D dxdy = ∫_(-2<=y<=1) dy ∫_(y-1<=x<=1-y^2) dx
= ∫_(-2<=y<=1) (1-y^2 - y +1) dy
= ∫_(-2<=y<=1) (2-y^2 - y) dy
=2(1+2) - (1/3) (1^3+2^3) - (1/2) (1-2^2)
=6 - 3 +3/2 = 9/2