volume bounded by two curves?

[匿名]

2016-04-10 7:28 am

Set up, bur do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line:

y=0, y=sinx for 0<=x<=pi about y=1

更新1:

(b) find the volume of the solid of revolution generated by revolving the region bounded by the curve y=2^(x^2) and the straight line y=2 about y-axis

回答 (1)

Lopez

2016-04-10 11:11 am

✔ 最佳答案

題目要求 "Set up, bur do not evaluate" , 也就是只列式, 不用後續求值.
以下兩小題皆可用 washer method.

(a)
因為 0 ≦ x ≦ π , 所以 0 ≦ sin x ≦ 1
R = ∣ 0 - 1 ∣ = 1
r = ∣ sin x - 1 ∣ = 1 - sin x

V
= ∫ π*( R^2 - r^2 ) dx , from x = 0 to x = π
= π * ∫ [ 1^2 - ( 1 - sin x )^2 ] dx , from x = 0 to x = π ..... Ans

(b)
先求兩函數之交點:
2^(x^2) = 2
x^2 = 1
x = ± 1
故積分區域為 [ - 1 , 1 ]
又 y = 2^(x^2) ≧ 1

R = ∣ 2 - 0 ∣ = 2
r = ∣ 2^(x^2) - 0 ∣ = 2^(x^2)

V
= ∫ π*( R^2 - r^2 ) dx , from x = - 1 to x = 1
= π * ∫ [ 2^2 - ( 2^(x^2) )^2 ] dx , from x = - 1 to x = 1 ..... Ans

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