次の曲線または直線で囲まれた図形をx軸のまわりに1回転してできる回転体の体積を求めます。 (1) $y = 3x$, $x = 1$, $x = 2$, $x$軸 (2) $y = \frac{1}{\sqrt{x}}$, $x = 1$, $x = 3$, $x$軸 (3) $y = e^{-x}$, $x = -1$, $x = 1$, $x$軸 (4) $y = \cos x$ ($0 \le x \le \frac{\pi}{2}$), $x$軸, $y$軸 (5) $y = \sqrt{3-x}$, $x$軸, $y$軸 (6) $y = \sqrt{1+x^2}$, $x = -1$, $x = 2$, $x$軸 (7) $y = \tan x$, $x = \frac{\pi}{4}$, $x$軸 (8) $y = \frac{1}{x+2}$, $x = 1$, $x$軸, $y$軸 (9) $\frac{x^2}{4} + \frac{y^2}{9} = 1$ (10) $y = 4-x^2$, $y = 3$ (11) $y = x^2 - x$, $x$軸 (12) $x^2 + y^2 + 2x = 0$

1. 問題の内容

次の曲線または直線で囲まれた図形をx軸のまわりに1回転してできる回転体の体積を求めます。

(1) $y = 3x$, $x = 1$, $x = 2$, $x$軸

(2) $y = \frac{1}{\sqrt{x}}$, $x = 1$, $x = 3$, $x$軸

(3) $y = e^{-x}$, $x = -1$, $x = 1$, $x$軸

(4) $y = \cos x$ ($0 \le x \le \frac{\pi}{2}$), $x$軸, $y$軸

(5) $y = \sqrt{3-x}$, $x$軸, $y$軸

(6) $y = \sqrt{1+x^2}$, $x = -1$, $x = 2$, $x$軸

(7) $y = \tan x$, $x = \frac{\pi}{4}$, $x$軸

(8) $y = \frac{1}{x+2}$, $x = 1$, $x$軸, $y$軸

(9) $\frac{x^2}{4} + \frac{y^2}{9} = 1$

(10) $y = 4-x^2$, $y = 3$

(11) $y = x^2 - x$, $x$軸

(12) $x^2 + y^2 + 2x = 0$

2. 解き方の手順

回転体の体積は、与えられた関数を積分して求めます。x軸周りの回転体の体積は、以下の公式で計算できます。

$V = \pi \int_a^b [f(x)]^2 dx$

ここで、$f(x)$は回転させる関数、$a$と$b$は積分範囲です。問題ごとに、積分範囲と関数を適切に設定して計算します。

(1)

$V = \pi \int_1^2 (3x)^2 dx = \pi \int_1^2 9x^2 dx = \pi [3x^3]_1^2 = \pi [3(2^3) - 3(1^3)] = \pi [24 - 3] = 21\pi$

(2)

$V = \pi \int_1^3 (\frac{1}{\sqrt{x}})^2 dx = \pi \int_1^3 \frac{1}{x} dx = \pi [\ln x]_1^3 = \pi (\ln 3 - \ln 1) = \pi \ln 3$

(3)

$V = \pi \int_{-1}^1 (e^{-x})^2 dx = \pi \int_{-1}^1 e^{-2x} dx = \pi [-\frac{1}{2}e^{-2x}]_{-1}^1 = \pi [-\frac{1}{2}e^{-2} + \frac{1}{2}e^2] = \frac{\pi}{2}(e^2 - e^{-2})$

(4)

$V = \pi \int_0^{\pi/2} (\cos x)^2 dx = \pi \int_0^{\pi/2} \frac{1 + \cos 2x}{2} dx = \pi [\frac{x}{2} + \frac{\sin 2x}{4}]_0^{\pi/2} = \pi [\frac{\pi}{4} + 0 - 0 - 0] = \frac{\pi^2}{4}$

(5)

$y = \sqrt{3-x}$と$x$軸、$y$軸で囲まれた図形なので、$x$の積分範囲は$0$から$3$まで。

$V = \pi \int_0^3 (\sqrt{3-x})^2 dx = \pi \int_0^3 (3-x) dx = \pi [3x - \frac{x^2}{2}]_0^3 = \pi [9 - \frac{9}{2} - 0] = \frac{9\pi}{2}$

(6)

$V = \pi \int_{-1}^2 (\sqrt{1+x^2})^2 dx = \pi \int_{-1}^2 (1+x^2) dx = \pi [x + \frac{x^3}{3}]_{-1}^2 = \pi [2 + \frac{8}{3} - (-1 - \frac{1}{3})] = \pi [2 + \frac{8}{3} + 1 + \frac{1}{3}] = \pi [3 + \frac{9}{3}] = 6\pi$

(7)

$V = \pi \int_0^{\pi/4} (\tan x)^2 dx = \pi \int_0^{\pi/4} (\frac{\sin^2 x}{\cos^2 x}) dx = \pi \int_0^{\pi/4} (\frac{1 - \cos^2 x}{\cos^2 x}) dx = \pi \int_0^{\pi/4} (\frac{1}{\cos^2 x} - 1) dx = \pi [\tan x - x]_0^{\pi/4} = \pi [1 - \frac{\pi}{4} - 0] = \pi (1 - \frac{\pi}{4})$

(8)

$V = \pi \int_0^1 (\frac{1}{x+2})^2 dx = \pi \int_0^1 (x+2)^{-2} dx = \pi [-(x+2)^{-1}]_0^1 = \pi [-\frac{1}{3} + \frac{1}{2}] = \frac{\pi}{6}$

(9)

$\frac{x^2}{4} + \frac{y^2}{9} = 1$を変形して、$y^2 = 9(1-\frac{x^2}{4}) = 9 - \frac{9}{4}x^2$

$V = \pi \int_{-2}^2 (9 - \frac{9}{4}x^2) dx = 2\pi \int_0^2 (9 - \frac{9}{4}x^2) dx = 2\pi [9x - \frac{3}{4}x^3]_0^2 = 2\pi [18 - \frac{3}{4}(8)] = 2\pi (18 - 6) = 24\pi$

(10)

$4 - x^2 = 3$より、$x^2 = 1$, $x = \pm 1$

$V = \pi \int_{-1}^1 (4-x^2)^2 - 3^2 dx = \pi \int_{-1}^1 (16 - 8x^2 + x^4 - 9) dx = \pi \int_{-1}^1 (7 - 8x^2 + x^4) dx = 2\pi \int_0^1 (7 - 8x^2 + x^4) dx = 2\pi [7x - \frac{8}{3}x^3 + \frac{x^5}{5}]_0^1 = 2\pi (7 - \frac{8}{3} + \frac{1}{5}) = 2\pi (\frac{105 - 40 + 3}{15}) = \frac{136\pi}{15}$

(11)

$V = \pi \int_0^1 (x^2 - x)^2 dx = \pi \int_0^1 (x^4 - 2x^3 + x^2) dx = \pi [\frac{x^5}{5} - \frac{x^4}{2} + \frac{x^3}{3}]_0^1 = \pi [\frac{1}{5} - \frac{1}{2} + \frac{1}{3}] = \pi [\frac{6 - 15 + 10}{30}] = \frac{\pi}{30}$

(12)

$x^2 + y^2 + 2x = 0$を変形して、$(x+1)^2 + y^2 = 1$。これは中心$(-1,0)$、半径1の円。

$V = \pi \int_{-2}^0 y^2 dx = \pi \int_{-2}^0 (-x^2 -2x) dx = \pi [-\frac{x^3}{3} - x^2]_{-2}^0 = \pi[0 - (\frac{8}{3} - 4)] = \pi [4 - \frac{8}{3}] = \frac{4\pi}{3}$

3. 最終的な答え

(1) $21\pi$

(2) $\pi \ln 3$

(3) $\frac{\pi}{2}(e^2 - e^{-2})$

(4) $\frac{\pi^2}{4}$

(5) $\frac{9\pi}{2}$

(6) $6\pi$

(7) $\pi (1 - \frac{\pi}{4})$

(8) $\frac{\pi}{6}$

(9) $24\pi$

(10) $\frac{136\pi}{15}$

(11) $\frac{\pi}{30}$

(12) $\frac{4\pi}{3}$