先求交点
联立
y²=ax
y=x²
解得
x=0,y=0
x=a^(1/3),y=a^(2/3)
∴S=∫(0,a^(1/3)) [√(ax)-x²]dx
=√a*∫(0,a^(1/3)) √xdx+∫(0,a^(1/3))x²dx
=(2/3)√a*[x^(3/2)]+(1/3)x³ |(0,a^(1/3))
=(2/3)a+(1/3)a
=a
∴a=9时,两曲线所围成的平面图形的面积为9
先求交点
联立
y²=ax
y=x²
解得
x=0,y=0
x=a^(1/3),y=a^(2/3)
∴S=∫(0,a^(1/3)) [√(ax)-x²]dx
=√a*∫(0,a^(1/3)) √xdx+∫(0,a^(1/3))x²dx
=(2/3)√a*[x^(3/2)]+(1/3)x³ |(0,a^(1/3))
=(2/3)a+(1/3)a
=a
∴a=9时,两曲线所围成的平面图形的面积为9