解法一（以x為積分變量求解）：
∵(自己作圖)x²+y²=2x與y=x的交點(diǎn)是(0,0)與(1,1)
∴所求面積=∫[√(2x-x²)-x]dx
=∫√(1-(x-1)²)dx-∫xdx
=∫cos²tdt-1/2 (在第一個積分中,令x-1=sint)
=∫[(1+cos(2t))/2]dt-1/2
=π/4-1/2
所求體積=∫2π(2-x)[√(2x-x²)-x]dx
=2π[∫(2-x)√(1-(x-1)²)dx-∫(2x-x²)dx
=2π[∫(1-sint)cos²tdt-(1-1/3)] (在第一個積分中,令x-1=sint)
=2π[∫(1/2+cos(2t)/2-sintcos²t)dt-2/3]
=2π[(1/3+π/4)-2/3]
=π²/2-2π/3
解法二（以y為積分變量求解）：
∵(自己作圖)x²+y²=2x與y=x的交點(diǎn)是(0,0)與(1,1)
∴所求面積=∫[y-(1-√(1-y²))]dy
=∫√(1-y²)dy+∫(y-1)dy
=∫cos²tdt+(1/2-1) (在第一個積分中,令y=sint)
=∫(1/2+cos(2t)/2)dt-1/2
=π/4-1/2
所求體積=∫π[(1+√(1-y²))²-(2-y)²]dy
=2π∫[√(1-y²)-(1-2y+y²)]dy
=2π[∫√(1-y²)dy-∫(1-2y+y²)dy]
=2π[∫cos²tdt-(1-1+1/3)] (在第一個積分中,令y=sint)
=2π[∫(1/2+cos(2t)/2)dt-1/3]
=2π(π/4-1/3)
=π²/2-2π/3