题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Figure 2 shows a sketch of part of the curve with equation
y=(2x2+3)1.512x
The region R, shown shaded in Figure 2, is bounded by the curve, the line with equation x=21, the x-axis and the line with equation x=k.
This region is rotated through 360∘ about the x-axis to form a solid of revolution.
Given that the volume of this solid is 648713π, use algebraic integration to find the exact value of the constant k.
(6)
题目中文翻译
本题中你必须写出解题过程的所有步骤。
完全依赖计算器技术的解法不被接受。
图 2 给出了部分曲线的示意图,其方程为
y=(2x2+3)1.512x
图 2 中阴影区域 R 由该曲线、直线 x=21、x 轴以及直线 x=k 围成。
该区域绕 x 轴旋转 360∘,形成一个旋转体。
已知该旋转体的体积为 648713π,用代数积分法求常数 k 的精确值。
解答
解法一
思路
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旋转体绕 x 轴生成,使用圆盘法 V=π∫y2dx。平方后,被积函数为 (2x2+3)3144x,分母内部的导数与分子成比例,可以直接积分。代入上下限并与已知体积相等,最后利用图中 k>0 选取正确的根。
答题过程
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The volume of the solid of revolution is
V=π∫21ky2dx.
Since
y2=(2x2+3)3144x,
we have
∫(2x2+3)3144xdx=−18(2x2+3)−2.
Therefore,
V===π[−(2x2+3)218]21kπ(−(2k2+3)218+4218)π(−(2k2+3)218+89).
Using the given volume,
−(2k2+3)218+89=648713.
Since 89=648729,
(2k2+3)218==64816812.
Hence
(2k2+3)2=729.
As 2k2+3>0,
2k2+3=k2=27,12.
The diagram shows that k is positive, so
k=23.