神马作文网求下列曲线所围成的图形的面积
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/23 20:46:28
求下列曲线所围成的图形的面积
x=a*cos^3(t)
y=a*sin^3(t)
x=a*cos^3(t)
y=a*sin^3(t)
该曲线为星形线,图形关于两坐标轴对称,因此下面只求第一象限,然后4倍就行了
S=4∫ [0-->a]ydx 将参数方程代入
=4∫ [π/2-->0]a*sin^3(t)*a*3cos^2(t)*(-sint)dt 整理并将上下限交换
=12a^2∫ [0-->π/2]sin^4(t)cos^2(t)dt
为了计算简便,下面用到一个结论
∫ [0-->π/2] f(sinx)dx=∫ [0-->π/2] f(cosx)dx （定积分的换元法部分的一道例题）
因此∫ [0-->π/2] f(sinx)dx=1/2[∫ [0-->π/2] f(sinx)dx+∫ [0-->π/2] f(cosx)dx]
则原式=12a^2∫ [0-->π/2]sin^4(t)cos^2(t)dt
=6a^2{∫ [0-->π/2]sin^4(t)cos^2(t)dt+∫ [0-->π/2]sin^2(t)cos^4(t)dt}
=6a^2∫ [0-->π/2]sin^2(t)cos^2(t)[sin^2(t)+cos^2(t)]dt
=6a^2∫ [0-->π/2]sin^2(t)cos^2(t)dt
=3/2*a^2∫ [0-->π/2]sin^2(2t)dt
=3/4*a^2∫ [0-->π/2] (1-cos4t)dt
=3/4*a^2[t-1/4sin4t] [0-->π/2]
=3a^2π/8
S=4∫ [0-->a]ydx 将参数方程代入
=4∫ [π/2-->0]a*sin^3(t)*a*3cos^2(t)*(-sint)dt 整理并将上下限交换
=12a^2∫ [0-->π/2]sin^4(t)cos^2(t)dt
为了计算简便,下面用到一个结论
∫ [0-->π/2] f(sinx)dx=∫ [0-->π/2] f(cosx)dx （定积分的换元法部分的一道例题）
因此∫ [0-->π/2] f(sinx)dx=1/2[∫ [0-->π/2] f(sinx)dx+∫ [0-->π/2] f(cosx)dx]
则原式=12a^2∫ [0-->π/2]sin^4(t)cos^2(t)dt
=6a^2{∫ [0-->π/2]sin^4(t)cos^2(t)dt+∫ [0-->π/2]sin^2(t)cos^4(t)dt}
=6a^2∫ [0-->π/2]sin^2(t)cos^2(t)[sin^2(t)+cos^2(t)]dt
=6a^2∫ [0-->π/2]sin^2(t)cos^2(t)dt
=3/2*a^2∫ [0-->π/2]sin^2(2t)dt
=3/4*a^2∫ [0-->π/2] (1-cos4t)dt
=3/4*a^2[t-1/4sin4t] [0-->π/2]
=3a^2π/8