求解一道 高数 二重积分 求面积的题
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y=x^2,y=1的交点坐标为A(-1,1),B(1,1),
是抛物线(开口向上),被平行于X轴的直线所截的区域,
-1<=x<=1,x^2<=y<=1,
I=∫ [-1,1]dx ∫ [x^2,1] √(y-x^2)dy
=(2/3)∫ [-1,1]dx ∫ (y-x^2)^(3/2)[x^2,1]
=(2/3)∫ [-1,1] (1-x^2)^(3/2)dx
设x=sint,dx=costdt,
=(2/3)∫ [-π/2,π/2](cost)^4dt
=(2/3)(1/4)∫[-π/2,π/2](1+cos2t^2dt
=(1/6)∫[-π/2,π/2][1+2cos2t+(cos2t)^2]dt
=(1/6)[t+sin2t+t/2+(sin4t)/4][ -π/2,π/2]
=π/4.
∴I=π/4
二重积分是求的体积.
是否还要求其面积?
是抛物线(开口向上),被平行于X轴的直线所截的区域,
-1<=x<=1,x^2<=y<=1,
I=∫ [-1,1]dx ∫ [x^2,1] √(y-x^2)dy
=(2/3)∫ [-1,1]dx ∫ (y-x^2)^(3/2)[x^2,1]
=(2/3)∫ [-1,1] (1-x^2)^(3/2)dx
设x=sint,dx=costdt,
=(2/3)∫ [-π/2,π/2](cost)^4dt
=(2/3)(1/4)∫[-π/2,π/2](1+cos2t^2dt
=(1/6)∫[-π/2,π/2][1+2cos2t+(cos2t)^2]dt
=(1/6)[t+sin2t+t/2+(sin4t)/4][ -π/2,π/2]
=π/4.
∴I=π/4
二重积分是求的体积.
是否还要求其面积?
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