Menentukan Rumus Volume Kerucut Terpancung dengan Menggunakan Integral
Apa itu Kerucut Terpancung?
Kerucut terpancung adalah kerucut yang bagian puncaknya terpotong mendatar. Sehingga secara sekilas seperti ember yang ditelungkupkan. Bangunnya seperti gambar berikut:
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzwBDlD9NJGyUFmwSLZKBz2Ugo9yhJbgJW-Lqm5f27z-UT2KqEEkiV8jwpX7JQUlCvGH0NkwJjbFkBoOOsKIOXEvoFx4CuEqCaPA9CO5V7AUY3cnjecujy8xjgC1HgeadSh1Fhar8MlBA/s320/kerucut-terpancung-2.webp)
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisNtgutTAU4fkQft3ny1lJLgosq1Z6DQ_K6Fwh4C4gCAJlA0j8cLFDzcypD6OUaBuuGO6UiyPwH1AG3kK2Ee_-aN8MN8wrCUsMj9sGKzs-ShJEAnPjRDVeHiwshcn1f7OzMmCDtqL1GcM/s320/kerucut-terpancung.webp)
VOLUME KERUCUT TERPANCUNG
Untuk mencari volume kerucut terpancung dengan rumus volume benda putar kita harus membuat grafik seperti gambar berikut
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgjBipPYonkYZ5TgoDJ28F-XhAUnYNIqIZTmQ9iZwKD4prnj4FyeArahVW8ZEjJ3QiIYgVKqG-LZAqTKxOrzP19HezYAe6ny7sQ7XiUmUkgIhpj46JIvuy0qlF_3DpTSfrWN2WQwz2hAIk/s400/kerucut-terpancung-3.webp)
Persamaan garis pada gambar di atas adalah
$\begin{array}{l}
\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}} \\
\Leftrightarrow \frac{{y - 0}}{{t - 0}} = \frac{{x - R}}{{r - R}} \\
\Leftrightarrow \frac{y}{t} = \frac{{x - R}}{{r - R}} \\
\Leftrightarrow t\left( {x - R} \right) = y\left( {r - R} \right) \\
\Leftrightarrow tx - tR = y\left( {r - R} \right) \\
\Leftrightarrow tx = y\left( {r - R} \right) + tR \\
\Leftrightarrow x = \frac{{y\left( {r - R} \right) + tR}}{t} \\
\Leftrightarrow x = \left( {\frac{{r - R}}{t}} \right)y + R \\
\end{array}$
\frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{x - {x_1}}}{{{x_2} - {x_1}}} \\
\Leftrightarrow \frac{{y - 0}}{{t - 0}} = \frac{{x - R}}{{r - R}} \\
\Leftrightarrow \frac{y}{t} = \frac{{x - R}}{{r - R}} \\
\Leftrightarrow t\left( {x - R} \right) = y\left( {r - R} \right) \\
\Leftrightarrow tx - tR = y\left( {r - R} \right) \\
\Leftrightarrow tx = y\left( {r - R} \right) + tR \\
\Leftrightarrow x = \frac{{y\left( {r - R} \right) + tR}}{t} \\
\Leftrightarrow x = \left( {\frac{{r - R}}{t}} \right)y + R \\
\end{array}$
Dengan menggunakan rumus volume benda putar, volume kerucut terpancung dapat dicari sebagai berikut:
$\begin{array}{l}
V = \pi \int\limits_0^t {{x^2}dy} \\
\Leftrightarrow V = \pi {\int\limits_0^t {\left[ {\left( {\frac{{r - R}}{t}} \right)y + R} \right]} ^2}dy \\
\Leftrightarrow V = \pi \int\limits_0^t {\left[ {{{\left( {\frac{{r - R}}{t}} \right)}^2}{y^2} + 2.R.\left( {\frac{{r - R}}{t}} \right)y + {R^2}} \right]} dy \\
\Leftrightarrow V = \pi \left[ {{{\left( {\frac{{r - R}}{t}} \right)}^2}.\frac{1}{3}{y^3} + 2R\left( {\frac{{r - R}}{t}} \right).\frac{1}{2}{y^2} + {R^2}y} \right]_0^t \\
\Leftrightarrow V = \pi \left[ {\left( {\frac{{{{\left( {r - R} \right)}^2}}}{{{t^2}}}.\frac{1}{3}{t^3} + R\left( {\frac{{r - R}}{t}} \right).{t^2} + {R^2}t} \right) - \left( 0 \right)} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}t{{\left( {r - R} \right)}^2} + Rt\left( {r - R} \right) + {R^2}t} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}t\left( {{r^2} - 2Rr + {R^2}} \right) + Rrt - {R^2}t + {R^2}t} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}{r^2}t - \frac{2}{3}Rrt + \frac{1}{3}{R^2}t + Rrt - {R^2}t + {R^2}t} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}{r^2}t + \frac{1}{3}Rrt + \frac{1}{3}{R^2}t} \right] \\
\Leftrightarrow V = \frac{1}{3}\pi t\left( {{r^2} + Rr + {R^2}} \right) \\
\end{array}$
V = \pi \int\limits_0^t {{x^2}dy} \\
\Leftrightarrow V = \pi {\int\limits_0^t {\left[ {\left( {\frac{{r - R}}{t}} \right)y + R} \right]} ^2}dy \\
\Leftrightarrow V = \pi \int\limits_0^t {\left[ {{{\left( {\frac{{r - R}}{t}} \right)}^2}{y^2} + 2.R.\left( {\frac{{r - R}}{t}} \right)y + {R^2}} \right]} dy \\
\Leftrightarrow V = \pi \left[ {{{\left( {\frac{{r - R}}{t}} \right)}^2}.\frac{1}{3}{y^3} + 2R\left( {\frac{{r - R}}{t}} \right).\frac{1}{2}{y^2} + {R^2}y} \right]_0^t \\
\Leftrightarrow V = \pi \left[ {\left( {\frac{{{{\left( {r - R} \right)}^2}}}{{{t^2}}}.\frac{1}{3}{t^3} + R\left( {\frac{{r - R}}{t}} \right).{t^2} + {R^2}t} \right) - \left( 0 \right)} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}t{{\left( {r - R} \right)}^2} + Rt\left( {r - R} \right) + {R^2}t} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}t\left( {{r^2} - 2Rr + {R^2}} \right) + Rrt - {R^2}t + {R^2}t} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}{r^2}t - \frac{2}{3}Rrt + \frac{1}{3}{R^2}t + Rrt - {R^2}t + {R^2}t} \right] \\
\Leftrightarrow V = \pi \left[ {\frac{1}{3}{r^2}t + \frac{1}{3}Rrt + \frac{1}{3}{R^2}t} \right] \\
\Leftrightarrow V = \frac{1}{3}\pi t\left( {{r^2} + Rr + {R^2}} \right) \\
\end{array}$
KESIMPULAN
Volume kerucut terpancung dengan jari-jari besar (R), jari-jari kecil (r) dan tinggi (t) adalah
\[V = \frac{1}{3}\pi t\left( {{r^2} + Rr + {R^2}} \right)\]
Demikian dan terima kasih.
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