求解一道微积分应用问题

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^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]