求抛物线y=1+((x)^(2))/4(0≤x≤2)绕x轴旋转所得的旋转体的表面积

Question

求抛物线y=1+((x)^(2))/4(0≤x≤2)绕x轴旋转所得的旋转体的表面积

Answer 1

图形绕x轴旋转所得的旋转体的表面积
=∫[2π(1+x²/4)√(1+y'²)]dx (y'=(1+x²/4)'=x/2)
=2π∫(1+x²/4)^(3/2)dx
=2π∫(sec³t*2sec²t)dt (令x/2=tant)
=4π∫[cost/(cost)^6]dt
=4π∫d(sint)/(1-sin²t)³
=4π∫(1/16)[2/(1-sint)³+3/(1-sint)²+3/(1-sint)+2/(1+sint)³+3/(1+sint)²+3/(1+sint)]d(sint)
=(π/4)[1/(1-sint)²+3/(1-sint)-3ln(1-sint)-1/(1+sint)²-3/(1+sint)+3ln(1+sint)]│
=(π/4)[4sint/(cost)^4+6sint/cos²t+3ln((1+sint)/(1-sint))]│
=(π/4)[4/(1/√2)³+6/(1/√2)+3ln((1+1/√2)/(1-1/√2))]
=(π/4)[8√2+6√2+6ln(√2+1)]
=(π/2)[7√2+3ln(√2+1)].