前置知识:定积分求旋转曲面的面积
习题
求曲线 y = 1 3 x 3 ( 0 ≤ x ≤ 1 ) y=\dfrac 13x^3(0\leq x\leq 1) y=31x3(0≤x≤1)绕 x x x轴旋转一周得到的旋转曲面的面积 S S S。
解:
S
=
2
π
∫
0
1
1
3
x
3
1
+
(
x
2
)
2
d
x
=
π
6
∫
0
1
1
+
x
4
d
(
1
+
x
4
)
\qquad S=2\pi\int_0^1\dfrac 13x^3\sqrt{1+(x^2)^2}dx=\dfrac{\pi}{6}\int_0^1\sqrt{1+x^4}d(1+x^4)
S=2π∫0131x31+(x2)2dx=6π∫011+x4d(1+x4)
= π 6 ⋅ [ 2 3 ( 1 + x 4 ) 3 2 ] ∣ 0 1 = π 9 ⋅ 2 3 2 − π 9 \qquad \quad =\dfrac{\pi}{6}\cdot[\dfrac 23(1+x^4)^{\frac 32}]\bigg\vert_0^1=\dfrac{\pi}{9}\cdot 2^{\frac 32}-\dfrac{\pi}{9} =6π⋅[32(1+x4)23] 01=9π⋅223−9π