【微分積分】4-7-4 回転体の体積と側面積|問題集
1.次の関数に囲まれる平面曲線を\(x\)軸の周りに回転してできる回転体の体積を求めなさい。
(1)\(y=x,y=0,x=1\)
回転体の体積\(V\)は
\(\displaystyle V=\pi\int_0^1x^2dx\)
\(\displaystyle =\pi\left[\frac{x^3}{3}\right]_0^1\)
\(\displaystyle =\frac{\pi}{3}\)
\(\displaystyle V=\pi\int_0^1x^2dx\)
\(\displaystyle =\pi\left[\frac{x^3}{3}\right]_0^1\)
\(\displaystyle =\frac{\pi}{3}\)
(2)\(y=x^2,y=9\)
回転体の体積\(V\)は
\(\displaystyle V=\pi\int_{-3}^3(9^2-(x^2)^2)dx\)
\(\displaystyle =2\pi\int_0^3(81-x^4)dx\)
\(\displaystyle =2\pi\left[81x-\frac{x^5}{5}\right]_0^3\)
\(\displaystyle =\frac{1944\pi}{5}\)
\(\displaystyle V=\pi\int_{-3}^3(9^2-(x^2)^2)dx\)
\(\displaystyle =2\pi\int_0^3(81-x^4)dx\)
\(\displaystyle =2\pi\left[81x-\frac{x^5}{5}\right]_0^3\)
\(\displaystyle =\frac{1944\pi}{5}\)
(3)\(y=\sqrt{x},y=x^3\)
回転体の体積\(V\)は
\(\displaystyle V=\pi\int_0^1\{(\sqrt{x})^2-(x^3)^2\}dx\)
\(\displaystyle =\pi\int_0^1(x-x^6)dx\)
\(\displaystyle =\pi\left[\frac{x^2}{2}-\frac{x^7}{7}\right]_0^1\)
\(\displaystyle =\frac{5\pi}{14}\)
\(\displaystyle V=\pi\int_0^1\{(\sqrt{x})^2-(x^3)^2\}dx\)
\(\displaystyle =\pi\int_0^1(x-x^6)dx\)
\(\displaystyle =\pi\left[\frac{x^2}{2}-\frac{x^7}{7}\right]_0^1\)
\(\displaystyle =\frac{5\pi}{14}\)
(4)\(y=x^2,y=x+2\)
回転体の体積\(V\)は
\(\displaystyle V=\pi\int_{-1}^2\{(x+2)^2-(x^2)^2\}dx\)
\(\displaystyle =\pi\int_{-1}^2(x^2+4x+4-x^4)dx\)
\(\displaystyle =\pi\left[\frac{x^3}{3}+2x^2+4x-\frac{x^5}{5}\right]_{-1}^2\)
\(\displaystyle =\frac{72\pi}{5}\)
\(\displaystyle V=\pi\int_{-1}^2\{(x+2)^2-(x^2)^2\}dx\)
\(\displaystyle =\pi\int_{-1}^2(x^2+4x+4-x^4)dx\)
\(\displaystyle =\pi\left[\frac{x^3}{3}+2x^2+4x-\frac{x^5}{5}\right]_{-1}^2\)
\(\displaystyle =\frac{72\pi}{5}\)
(5)\(x^2+(y-2)^2\leqq1\)
回転体の体積\(V\)は
\(\displaystyle V=\pi\int_{-1}^1\{(2+\sqrt{1-x^2})^2-(2-\sqrt{1-x^2})^2\}dx\)
\(\displaystyle =\pi\int_{-1}^18\sqrt{1-x^2}dx\)
\(\displaystyle =8\pi・\frac{\pi}{2}\)
\(\displaystyle =4\pi^2\)
\(\displaystyle V=\pi\int_{-1}^1\{(2+\sqrt{1-x^2})^2-(2-\sqrt{1-x^2})^2\}dx\)
\(\displaystyle =\pi\int_{-1}^18\sqrt{1-x^2}dx\)
\(\displaystyle =8\pi・\frac{\pi}{2}\)
\(\displaystyle =4\pi^2\)
(6)\(x=t-\sin t,y=1-\cos t,0\leqq t\leqq 2\pi\)
回転体の体積\(V\)は
\(\displaystyle V=\int_0^{2\pi}\pi y^2dx\)
\(\displaystyle =2\int_0^\pi\pi y^2dx\)
\(\displaystyle =2\pi\int_0^\pi(1-\cos t)^3dt\)
\(\displaystyle =16\pi\int_0^\pi\sin^6\frac{t}{2}dt\)
\(\displaystyle =16\pi・\frac{5!!}{6!!}・\frac{\pi}{2}\)
\(\displaystyle =5\pi^2\)
\(\displaystyle V=\int_0^{2\pi}\pi y^2dx\)
\(\displaystyle =2\int_0^\pi\pi y^2dx\)
\(\displaystyle =2\pi\int_0^\pi(1-\cos t)^3dt\)
\(\displaystyle =16\pi\int_0^\pi\sin^6\frac{t}{2}dt\)
\(\displaystyle =16\pi・\frac{5!!}{6!!}・\frac{\pi}{2}\)
\(\displaystyle =5\pi^2\)
次の学習に進もう!