用极坐标,∫∫e^(-r^2) rdrdθ=π[1-1/e]=π(e-1)/e>π/2∫∫sin(r^2) rdrdθ=π(1-cos1)≈π(1-1+1/2)=π/2∫∫cos(r^2) rdrdθ=πsin1≈π(1-1/6)=5π/6所以 I3>I1>I2第十题,很明显,越往外,x^2+y^2越大,1-x^2-y^2越小,所以I1>I2>I3
