【高校数学Ⅲ】6-1-2 体積｜問題集

正三角形\(PQR\)の面積を\(S(x)\)とすると、
\(\displaystyle S(x)=\frac{1}{2}\cos x・\frac{\sqrt{3}}{2}\cos x\)
\(\displaystyle =\frac{\sqrt{3}}{4}\cos^2x\)
よって、求める体積\(V\)は
\(\displaystyle V=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} S(x)dx\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sqrt{3}}{4}\cos^2x\)
\(\displaystyle =\frac{\sqrt{3}}{4}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos2x}{2}dx\)
\(\displaystyle =\frac{\sqrt{3}}{8}\left[x+\frac{1}{2}\sin2x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\)
\(\displaystyle =\frac{\sqrt{3}}{8}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\)
\(\displaystyle =\frac{\sqrt{3}}{8}\pi\)

断面の高さを\(x\)とすると、
断面:底面\(=x:4\)
断面積:底面積面\(=x^2:4^2\)
断面積\(S(x)\)は
\(\displaystyle S(x)=\frac{25x^2}{16}\)
よって、求める体積\(V\)は
\(\displaystyle V=\int_0^4 S(x)dx\)
\(\displaystyle =\frac{25}{16}\int_0^4 x^2dx\)
\(\displaystyle =\frac{25}{48}[x^3]_0^4\)
\(\displaystyle =\frac{25}{48}・4^3\)
\(\displaystyle =\frac{100}{3}\)

直円柱の高さを\(h\)とすると、
\(h=\sqrt{3}\sqrt{a^2-x^2}\)
断面積\(S(x)\)は
\(\displaystyle S(x)=\frac{\sqrt{3}}{2}\sqrt{a^2-x^2}\)
よって、求める体積\(V\)は
\(\displaystyle V=\int_{-a}^a S(x)dx\)
\(\displaystyle =\frac{\sqrt{3}}{2}\int_{-a}^a (a^2-x^2)dx\)
\(\displaystyle =\frac{\sqrt{3}}{2}\left[ax^2-\frac{1}{3}x^3\right]_{-a}^a\)
\(\displaystyle =\frac{\sqrt{3}}{2}\left(a^3-\frac{1}{3}a^3+a^3-\frac{1}{3}a^3\right)\)
\(\displaystyle =\frac{2\sqrt{3}}{3}a^3\)

\(y=2-e^x\)と\(x\)軸との交点は
\(\displaystyle 2-e^x=0\)
\(e^x=2\)
\(x=\log2\)
求める体積\(V\)は
\(\displaystyle V=\pi\int_0^{\log2} y^2dx\)
\(\displaystyle =\pi\int_0^{\log2} (2-e^x)^2dx\)
\(\displaystyle =\pi\int_0^{\log2} (4-4e^x+e^{2x})dx\)
\(\displaystyle =\pi\left[4x-4e^x+\frac{1}{2}e^{2\log2}\right]_0^{\log2}\)
\(\displaystyle =\pi\left(4\log2-4・2+\frac{1}{2}・2^2+\frac{7}{2}\right)\)
\(\displaystyle =\left(4\log2-\frac{5}{2}\right)\pi\)

\(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)より、\(\displaystyle y^2=\frac{b^2}{a^2}(a^2-x^2)\)
求める体積\(V\)は
\(\displaystyle V=\pi\int_{-a}^a y^2dx\)
\(\displaystyle =\frac{\pi b^2}{a^2}\int_{-a}^a (a^2-x^2)dx\)
\(\displaystyle =\frac{\pi b^2}{a^2}・\frac{4}{3}a^3\)
\(\displaystyle =\frac{4}{3}ab^2\pi\)

\(y=4x-x^2\)と\(y=x\)との交点は
\(4x-x^2=x\)
\(x^2-3x=0\)
\(x(x-3)=0\)
\(x=0,3\)
\(0\leqq x\leqq3\)のとき、\(4x-x^2\geqq x\)
よって、求める体積\(V\)は
\(\displaystyle V=\pi\int_0^3 (4x-x^2)^2dx-\pi\int_0^3 x^2dx\)
\(\displaystyle =\pi\int_0^3 (15x^2-8x^3+x^4)dx\)
\(\displaystyle =\pi\left[5x^3-2x^4+\frac{1}{5}x^5\right]_0^3\)
\(\displaystyle =\frac{108}{5}\pi\)

\(y=\sqrt{x}\)と\(y=x\)との交点は
\(\sqrt{x}=x\)
\(x^2-x=0\)
\(x(x-1)=0\)
\(x=0,1\)
\(0\leqq x\leqq1\)のとき、\(\sqrt{x}\geqq x\)
よって、求める体積\(V\)は
\(\displaystyle V=\pi\int_0^1 (\sqrt{x})^2dx-\pi\int_0^1 x^2dx\)
\(\displaystyle =\pi\int_0^1 (x-x^2)dx\)
\(\displaystyle =\pi\left[\frac{1}{2}x^2-\frac{1}{3}x^3\right]_0^1\)
\(\displaystyle =\frac{1}{6}\pi\)

求める体積\(V\)は
\(\displaystyle V=\pi\int_1^4 x^2dy\)
\(\displaystyle =\pi\int_1^4 (4-y)dy\)
\(\displaystyle =\pi\left[4y-\frac{1}{2}y^2\right]_1^4\)
\(\displaystyle =\pi\left(16-8-4+\frac{1}{2}\right)\)
\(\displaystyle =\frac{9}{2}\pi\)

\(y=1-\sqrt{x}\)より、\(x^2=(1-y)^4\)
求める体積\(V\)は
\(\displaystyle V=\pi\int_0^1 x^2dy\)
\(\displaystyle =\pi\int_0^1 (1-y)^4dy\)
\(\displaystyle =-\frac{1}{5}\pi[(1-y)^5]_0^1\)
\(\displaystyle =\frac{1}{5}\pi\)

\(\displaystyle \frac{dx}{d\theta}=3\cos^2\theta-\sin\theta\)
\(dx=-3\sin\theta\cos^2\theta d\theta\)
求める体積\(V\)は
\(\displaystyle V=2\pi\int_0^1 y^2dx\)
\(\displaystyle =6\pi\int_0^{\frac{\pi}{2}} \sin^7\theta\cos^2\theta d\theta\)
\(\displaystyle =6\pi\int_0^{\frac{\pi}{2}} \sin^7\theta(1-\sin^2\theta) d\theta\)
\(\displaystyle =6\pi\int_0^{\frac{\pi}{2}} (\sin^7\theta-\sin^9\theta) d\theta\)
\(\displaystyle =6\pi\left(\frac{6・4・2}{7・5・3}-\frac{8・6・4・2}{9・7・5・3}\right)\)
\(\displaystyle =\frac{48}{105}\pi\)