以x^2+y^2=r^2为例:4∫[0～r]√(r^2-x^2)dx上式可用换元法发来算,我以为你会呢,所以没写,!设:x=rsint则上式变为4∫[0～π/2]rcostd(rsint)=4∫[0～π/2]r^2(cost)^2dt=4r^2∫[0～π/2](1+cos2t)/2dt=4r^2(π/4+∫[0...