Question

求解一道微积分应用问题

求两个椭圆x^2/a^2+y^2/b^2=1和x^2/b^2+y^2/a^2=1相交部分的面积

谢谢大家了！！

Answer 1

用极坐标算
x^2/a^2+y^2/b^2=1
设
x=acost
y=bsint
x^2/b^2+y^2/a^2=1
x=bcost
y=asint
两方程联立
x^2/a^2+y^2/b^2=1
x^2/b^2+y^2/a^2=1
解得
x=+-y=+-ab/根号（a^2+b^2）
S1=∫（-π/4，π/4）1/2*R^2dt
=∫（-π/4，π/4）1/2*(a^2cos^t+b^2sin^t)dt
=1/2*∫（-π/4，π/4）(a^2+(b^2-a^2)sin^t)dt
=1/2*(a^2*π/2+(b^2-a^2)*(1/2-sin2t /4(t从-π/4，π/4)))
=1/2*[（2+π/2）*a^2+2b^2]
S=4*S1
=2*[（2+π/2）*a^2+2b^2]