📚 Example 6.3.2 | 例题6.3.2
This worked example explores a classic A-Level Mathematics problem that requires combining differentiation, curve sketching, and definite integration. We will walk through every step in detail, demonstrating the logical thinking expected in the AQA examination.
本例题探讨一个经典的A-Level数学问题，它要求将微分、曲线作图以及定积分结合起来。我们将详细地逐步推导，展示AQA考试所期待的严谨逻辑思维。
1. Problem Statement | 题目陈述
A curve is defined by the equation y = x²(1 − x). The region bounded by the curve, the x-axis, and the vertical lines x = 0 and x = 1 is rotated through 2π radians about the x-axis. Find the exact volume of the solid of revolution formed.
一条曲线由方程 y = x²(1 − x) 定义。由该曲线、x轴以及两条垂直线 x = 0 和 x = 1 所围成的区域，绕x轴旋转 2π 弧度。求所形成旋转体的精确体积。
Before jumping into the integration, let us first understand the behaviour of the function and confirm the region we are working with.
在开始积分计算之前，让我们首先理解这个函数的性质，并确认我们所要处理的区域。
2. Expanding and Simplifying | 展开与化简
We begin by expanding the given function to make subsequent differentiation and integration more straightforward. The equation y = x²(1 − x) expands to y = x² − x³.
我们首先展开给定的函数，以简化后续的微分和积分运算。方程 y = x²(1 − x) 展开后得到 y = x² − x³。
y = x² − x³
The expanded form reveals that this is a cubic polynomial with no constant term. This means the curve passes through the origin, which will be important when we verify the region.
展开后的形式表明这是一个没有常数项的三次多项式。这意味着曲线经过原点，这一点在验证区域时非常重要。
3. Roots and Intercepts | 根与截距
To sketch the curve and understand the region bounded by it, we must first determine the x-intercepts. Setting y = 0 yields x²(1 − x) = 0, which has solutions x = 0 and x = 1.
为了绘制曲线草图并理解它所围成的区域，我们必须首先确定x轴截距。令 y = 0，得到 x²(1 − x) = 0，其解为 x = 0 和 x = 1。
| Factor | Root | Factor | 根 |
| x² = 0 | x = 0 | x² = 0 | x = 0 |
| 1 − x = 0 | x = 1 | 1 − x = 0 | x = 1 |
Thus the curve intersects the x-axis at the origin and at the point (1, 0). Since the vertical lines in the problem are x = 0 and x = 1, the region is clearly contained between these two intercepts.
因此，曲线在原点以及点 (1, 0) 处与x轴相交。由于题目中的垂直线是 x = 0 和 x = 1，该区域显然位于这两个截距之间。
4. First Derivative and Critical Points | 一阶导数与临界点
To understand the shape of the curve, we differentiate y with respect to x. The derivative of y = x² − x³ is dy/dx = 2x − 3x².
为了理解曲线的形状，我们对y关于x进行微分。y = x² − x³ 的导数为 dy/dx = 2x − 3x²。
dy/dx = 2x − 3x² = x(2 − 3x)
Setting the derivative equal to zero gives the stationary points: x = 0 and x = 2/3. The second derivative, d²y/dx² = 2 − 6x, helps us classify these points.
令导数为零，可得驻点：x = 0 和 x = 2/3。二阶导数 d²y/dx² = 2 − 6x 有助于我们对这些点进行分类。
5. Nature of the Stationary Point | 驻点的性质
At x = 2/3, the second derivative evaluates to 2 − 6(2/3) = 2 − 4 = −2, which is negative. Therefore, x = 2/3 corresponds to a local maximum.
在 x = 2/3 处，二阶导数的值为 2 − 6(2/3) = 2 − 4 = −2，为负数。因此，x = 2/3 对应一个局部极大值。
The y-value at this maximum is found by substituting x = 2/3 back into the original equation:
该极大值处的y值，可通过将 x = 2/3 代回原方程求得：
y = (2/3)²(1 − 2/3) = (4/9)(1/3) = 4/27
This confirms the maximum point is at (2/3, 4/27). For x = 0, the second derivative is positive, so the origin is a local minimum, though it lies on the x-axis.
这证实极大值点为 (2/3, 4/27)。对于 x = 0，二阶导数为正，因此原点是局部极小值，尽管它位于x轴上。
6. Region and Curve Analysis | 区域与曲线分析
On the interval [0, 1], we observe that y = x²(1 − x) ≥ 0. Since x² is always non-negative and (1 − x) is non-negative for x ≤ 1, the curve lies entirely on or above the x-axis throughout this interval.
在区间 [0, 1] 上，我们观察到 y = x²(1 − x) ≥ 0。由于 x² 始终非负，且当 x ≤ 1 时 (1 − x) 非负，因此曲线在该区间内完全位于x轴上或上方。
The region is therefore a simple closed area bounded by the arc of the curve from x = 0 to x = 1 and the x-axis segment between those same limits. This confirms the rotational volume formula applies without modification.
因此，该区域是一个简单封闭区域，由 x = 0 到 x = 1 的曲线弧以及x轴上的相应线段围成。这证实了旋转体体积公式可以直接应用，无需修正。
7. Setting Up the Volume Integral | 建立体积积分
When a region under a curve y = f(x) is rotated about the x-axis, the volume of the solid formed is given by the formula V = π∫ₐᵇ y² dx, where a and b are the limits of integration.
当曲线 y = f(x) 下方的区域绕x轴旋转时，所形成旋转体的体积由公式 V = π∫ₐᵇ y² dx 给出，其中 a 和 b 是积分的上下限。
V = π∫₀¹ [x²(1 − x)]² dx
Here a = 0 and b = 1, since these are the vertical lines specified in the problem and they coincide with the x-intercepts of the curve.
这里 a = 0，b = 1，因为这是题目指定的垂直线，并且它们与曲线的x轴截距一致。
8. Simplifying the Integrand | 化简被积函数
We must square the original function carefully. The expression [x²(1 − x)]² can be written as x⁴(1 − x)², which expands to x⁴(1 − 2x + x²).
我们必须仔细地平方原始函数。表达式 [x²(1 − x)]² 可以写成 x⁴(1 − x)²，展开后为 x⁴(1 − 2x + x²)。
y² = x⁴ − 2x⁵ + x⁶
Expanding the terms gives y² = x⁴ − 2x⁵ + x⁶. This polynomial form is much easier to integrate term by term.
展开各项后得到 y² = x⁴ − 2x⁵ + x⁶。这个多项式形式更容易逐项积分。
9. Performing the Integration | 执行积分
We now integrate each term with respect to x. The integral of x⁴ is x⁵/5, the integral of −2x⁵ is −2x⁶/6 = −x⁶/3, and the integral of x⁶ is x⁷/7.
我们现在对每一项关于x进行积分。x⁴ 的积分是 x⁵/5，−2x⁵ 的积分是 −2x⁶/6 = −x⁶/3，x⁶ 的积分是 x⁷/7。
∫₀¹ (x⁴ − 2x⁵ + x⁶) dx = [x⁵/5 − x⁶/3 + x⁷/7]₀¹
It is always good practice to simplify the coefficients before substituting the limits, which helps avoid arithmetic errors.
在代入上下限之前，先化简系数是一个好的习惯，这有助于避免算术错误。
10. Evaluating the Definite Integral | 计算定积分
Substituting x = 1 into the antiderivative gives 1/5 − 1/3 + 1/7. Substituting x = 0 gives 0, since every term contains a power of x and therefore vanishes at the lower limit.
将 x = 1 代入原函数得到 1/5 − 1/3 + 1/7。将 x = 0 代入得到 0，因为每一项都含有x的幂，因此在下限处全部消失。
To combine these fractions, we find the least common denominator. The denominators are 5, 3, and 7, whose LCM is 105. Thus:
为了合并这些分数，我们找到最小公分母。分母为 5、3、7，它们的最小公倍数为 105。因此：
1/5 = 21/105, 1/3 = 35/105, 1/7 = 15/105
Combining these yields (21 − 35 + 15)/105 = 1/105. Thus the definite integral evaluates to 1/105.
将它们合并得到 (21 − 35 + 15)/105 = 1/105。因此，定积分的值为 1/105。
11. Final Volume and Interpretation | 最终体积与解读
Multiplying the result of the integral by π gives the exact volume of the solid of revolution:
将积分结果乘以π，即可得到旋转体的精确体积：
V = π × (1/105) = π/105
The exact volume is therefore π/105 cubic units. This is a relatively small volume, which is consistent with the fact that the maximum y-value of the curve is only 4/27 ≈ 0.148, so the rotated cross-sections are quite small throughout.
因此，精确体积为 π/105 立方单位。这是一个相对较小的体积，这与曲线的最大y值仅为 4/27 ≈ 0.148 这一事实相符，因此整个旋转过程中的横截面积都非常小。
12. Common Mistakes and Tips | 常见错误与技巧
One common mistake is forgetting to square the function before integrating when using the volume formula. Students sometimes integrate y directly instead of y², which yield a completely different and incorrect result.
一个常见错误是在使用体积公式时，忘记在积分前对函数进行平方。学生有时直接对 y 积分而不是对 y² 积分，这会得出完全不同的错误结果。
Another frequent error is mishandling the expansion of [x²(1 − x)]². Expanding it as x²(1 − x)² rather than x⁴(1 − x)² changes the degree of the polynomial and leads to an incorrect integral.
另一个常见错误是处理 [x²(1 − x)]² 的展开。将其展开为 x²(1 − x)² 而不是 x⁴(1 − x)²，会改变多项式的次数，从而导致错误的积分。
The key tip for this type of question is to always verify the region and the sign of the function before applying the volume formula. If the curve dips below the x-axis, additional care is needed.
解决这类问题的关键技巧是，在应用体积公式之前，始终验证区域以及函数的符号。如果曲线下降到x轴以下，则需要格外小心。
Finally, when calculating exact values, keep fractions throughout and use the lowest common denominator when combining them. Converting to decimals prematurely can introduce rounding errors that may cost marks.
最后，在计算精确值时，全程保留分数形式，并在合并分数时使用最小公分母。过早地转换为小数可能会引入四舍五入误差，从而失分。
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