文档介绍：第八节定积分的几何应用举例一、平面图形的面积二、体积三、平面曲线的弧长一、平面图形的面积1、直角坐标系情形设曲线y=f (x)(x?0) 与直线x= a , x = b (a <b) 及x轴所围曲边梯形的面积为A , 则xyo)(xfy?abxxxd?,d)(dxxfA?.d)(??baxxfA如右下图所示图形的面积:xyo)(1xfy?)(2xfy?abxxxd?,d)]()([d12xxfxfA??.d)]()([12???baxxfxfA如图所示图形面积为xxfxfAbad|)()(|21???yobxa)(2xfy?)(1xfy?xxxd?解xxy?2oy2xy?例1计算由两条抛物线y2 = x 和y=x2所围成的图形的面积.?????22xyxy由得两曲线交点,)1,1(,)0,0(xxxd?)1,1(1面积元素,d)(d2xxxA??xxxAd)(210????10333223?????????问题:积分变量只能选x吗?xyo)(yx??cd曲边梯形的面积yyyd???dcyyAd)(?yyAd)(d??)(1yx??)(2yx??xyocdyyyd?图形的面积yyyAd)]()([d12???????dcyyyAd)]()([12??解例1计算由两条抛物线y2 = x 和y=x2所围成的图形的面积.?????22xyxy由得两曲线交点,)1,1(,)0,0(x2yx?oyyx?)1,1(面积元素,d)(d2yyyA??xyyAd)(210????10333223?????????解题步骤:1. . 写出微元(面积元素)dA .2. . ??baAA3. 确定一个积分变量及其变化区间[a , b ] .例2计算由曲线xy22?和直线4??).4,8(),2,2(???????422xyxy由xy22?4??xy例2计算由曲线xy22?和直线4??).4,8(),2,2(???????422xyxy由xy22?4??xyyyyd?,d)24(d2yyyA??????????422d)24(yyyA423261421???????????例3计算由曲线xxy63??和2xy??xxy63??21AAA??1A2A解得交点为.)9,3(,)4,2(,)0,0(?????????236xyxxy由],0,2[,d)6(d231?????xxxxxA],3,0[,d)6(d322????xxxxxAxxxxd)6(3230????.12253???说明:)6(2023??????