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CIE 9709 2025 Nov Paper 11 Q8

A Level / CIE / P1

CIE 9709 2025 Nov Paper 11 Paper · Question 8

题目

Problem

The diagram shows the curve with equation y=12xy=\frac{1}{2}\sqrt{x} and the point PP with coordinates (9,32)\left(9,\frac{3}{2}\right). The shaded region is bounded by the curve and the lines x=0x=0 and y=32y=\frac{3}{2}.

(a) Find the area of the shaded region.

[3]

(b) The shaded region is rotated through 360∘360^\circ about the yy-axis.

Find the exact volume of the solid produced.

[4]
题目中文翻译

图中显示曲线 y=12xy=\frac{1}{2}\sqrt{x},以及坐标为 (9,32)\left(9,\frac{3}{2}\right) 的点 PP。阴影区域由曲线以及直线 x=0x=0 和 y=32y=\frac{3}{2} 围成。

(a) 求阴影区域的面积。

(b) 将阴影区域绕 yy 轴旋转 360∘360^\circ。

求所得立体的精确体积。

解答

(a)

解法一

思路

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竖直切分时,阴影在水平线与曲线之间。从整个长方形面积中减去曲线下面的面积。

答题过程

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The horizontal boundary is at height 32\frac32 from x=0x=0 to x=9x=9. The shaded area is

 32(9)−∫0912x1/2 dx= 272−[13x3/2]09= 272−9=92.\begin{align*} &\,\frac32(9)-\int_0^9\frac12x^{1/2}\,\mathrm{d}x\\ =&\,\frac{27}2-\Big[\frac13x^{3/2}\Big]_0^9\\ =&\,\frac{27}2-9=\frac92. \end{align*}

解法二

思路

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也可以水平切分。把 xx 表成 yy 的函数后,左边界 x=0x=0 到曲线的距离就是每条横条的长度。

答题过程

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The curve y=12xy=\frac12\sqrt x gives x=4y2x=4y^2 for y≥0y\ge0. The horizontal width of the region is 4y24y^2, so the area is

∫03/24y2 dy=[43y3]03/2=92.\int_0^{3/2}4y^2\,\mathrm{d}y =\Big[\frac43y^3\Big]_0^{3/2}=\frac92.

(b)

解法一

思路

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绕 yy 轴旋转后,水平截面是圆盘。半径等于曲线的横坐标 x=4y2x=4y^2。

答题过程

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At height yy, the disc has radius 4y24y^2, so its area is 16πy416\pi y^4. The volume is

V=16π∫03/2y4 dy=16π5[y5]03/2=243π10.V=16\pi\int_0^{3/2}y^4\,\mathrm{d}y =\frac{16\pi}{5}\Big[ y^5\Big]_0^{3/2} =\frac{243\pi}{10}.