令x=r*siny,0<=y<=π/2
原积分式=（上π/2下0）∫√(r^2－x^2)dx=（上π/2下0）(r^2)*∫(cosy)^2dy
=（上π/2下0）r*∫(cosy)^2dy==（上π/2下0）r*∫(cos2y+1)/2dy
可解得(r^2)*∫(cos2y+1)/2dy=1/4(sin2y+2y)*(r^2)
所以有
（上r下0）∫√(r^2－x^2)dx=1/4(sinπ+π)*(r^2)-1/4(sin0+0)*(r^2)=(r^2)π/4
结果是(r^2)π/4