题目
Problem
The diagram shows the curve with equation y = 1 2 x y=\frac{1}{2}\sqrt{x} y = 2 1 x and the point P P P with coordinates ( 9 , 3 2 ) \left(9,\frac{3}{2}\right) ( 9 , 2 3 ) . The shaded region is bounded by the curve and the lines x = 0 x=0 x = 0 and y = 3 2 y=\frac{3}{2} y = 2 3 .
(a) Find the area of the shaded region.
[3]
(b) The shaded region is rotated through 360 ∘ 360^\circ 36 0 ∘ about the y y y -axis.
Find the exact volume of the solid produced.
[4]
题目中文翻译
图中显示曲线 y = 1 2 x y=\frac{1}{2}\sqrt{x} y = 2 1 x ,以及坐标为 ( 9 , 3 2 ) \left(9,\frac{3}{2}\right) ( 9 , 2 3 ) 的点 P P P 。阴影区域由曲线以及直线 x = 0 x=0 x = 0 和 y = 3 2 y=\frac{3}{2} y = 2 3 围成。
(a) 求阴影区域的面积。
(b) 将阴影区域绕 y y y 轴旋转 360 ∘ 360^\circ 36 0 ∘ 。
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。
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The horizontal boundary is at height 3 2 \frac32 2 3 from x = 0 x=0 x = 0 to x = 9 x=9 x = 9 . The shaded area is
3 2 ( 9 ) − ∫ 0 9 1 2 x 1 / 2 d x = 27 2 − [ 1 3 x 3 / 2 ] 0 9 = 27 2 − 9 = 9 2 . \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*} = = 2 3 ( 9 ) − ∫ 0 9 2 1 x 1/2 d x 2 27 − [ 3 1 x 3/2 ] 0 9 2 27 − 9 = 2 9 .
解法二
思路
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也可以水平切分。把 x x x 表成 y y y 的函数后,左边界 x = 0 x=0 x = 0 到曲线的距离就是每条横条的长度。
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The curve y = 1 2 x y=\frac12\sqrt x y = 2 1 x gives x = 4 y 2 x=4y^2 x = 4 y 2 for y ≥ 0 y\ge0 y ≥ 0 . The horizontal width of the region is 4 y 2 4y^2 4 y 2 , so the area is
∫ 0 3 / 2 4 y 2 d y = [ 4 3 y 3 ] 0 3 / 2 = 9 2 . \int_0^{3/2}4y^2\,\mathrm{d}y
=\Big[\frac43y^3\Big]_0^{3/2}=\frac92. ∫ 0 3/2 4 y 2 d y = [ 3 4 y 3 ] 0 3/2 = 2 9 .
(b)
解法一
思路
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绕 y y y 轴旋转后,水平截面是圆盘。半径等于曲线的横坐标 x = 4 y 2 x=4y^2 x = 4 y 2 。
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At height y y y , the disc has radius 4 y 2 4y^2 4 y 2 , so its area is 16 π y 4 16\pi y^4 16 π y 4 . The volume is
V = 16 π ∫ 0 3 / 2 y 4 d y = 16 π 5 [ y 5 ] 0 3 / 2 = 243 π 10 . 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}. V = 16 π ∫ 0 3/2 y 4 d y = 5 16 π [ y 5 ] 0 3/2 = 10 243 π .