用极坐标算x^2/a^2+y^2/b^2=1设x=acosty=bsintx^2/b^2+y^2/a^2=1x=bcosty=asint两方程联立x^2/a^2+y^2/b^2=1x^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^...