Mencari Rumus Volume Bangun Ruang Torus (Bangun Mirip Donat/Ban) dengan Integral

Mungkin banyak yang belum tahu apa itu TORUS ya? Dalam kehidupan nyata torus itu bangun yang mirip donat, ban sepeda atau mobil. Pada bangku sekolah, bangun ini juga sepertinya kurang mendapat perhatian sehingga banyak yang belum tahu.

Secara geometris bangun torus seperti gambar berikut

Terdapat dua jari-jari pada torus yaitu jari-jari besar (R) dan jari-jari kecil (r).

VOLUME TORUS

Sama seperti postingan sebelumnya, pada kali ini pun saya akan mencoba mencari volume bangun Torus dengan menggunakan volume benda putar. untuk bangun torus, gambarnya sebagai berikut:

Lingkaran dengan diameter $\left( {R - r} \right)$ dengan titik pusat $\left( {0,\frac{1}{2}\left( {R + r} \right)} \right)$ memotong sumbu $Y$ di titik $\left( {0,R} \right)$ dan $\left( {0,r} \right)$ diputar sebesar ${360^o}$ terhadap sumbu $X$. Pesamaan lingkaran tersebut adalah sebagai berikut:

$\begin{array}{l}{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}\\ \Leftrightarrow {\left( {x - 0} \right)^2} + {\left( {y - \frac{1}{2}\left( {R + r} \right)} \right)^2} = {\left( {\frac{1}{2}\left( {R - r} \right)} \right)^2}\\ \Leftrightarrow {x^2} + {\left( {y - \frac{1}{2}\left( {R + r} \right)} \right)^2} = {\left( {\frac{1}{2}\left( {R - r} \right)} \right)^2}\\ \Leftrightarrow {\left( {y - \frac{1}{2}\left( {R + r} \right)} \right)^2} = {\left( {\frac{1}{2}\left( {R - r} \right)} \right)^2} - {x^2}\\ \Leftrightarrow y - \frac{1}{2}\left( {R + r} \right) = \sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}} \\ \Leftrightarrow y = \frac{1}{2}\left( {R + r} \right) \pm \sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}} \\ \Leftrightarrow {y_1} = \frac{1}{2}\left( {R + r} \right) + \sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}}  \vee {y_2} = \frac{1}{2}\left( {R + r} \right) - \sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}} \end{array}$

Dari hasil terakhir terdapat 2 persamaan bola yaitu yang berwarna merah untuk ${y_1}$ dan biru untuk ${y_1}$. Dari persamaan lingkaran di atas, kita gunakan untuk mencari volume benda putar untuk dua fungsi yaitu sebagai berikut:

$\begin{array}{l}V = \pi \int\limits_a^b {\left( {y_1^2 - y_2^2} \right)dx} \\ \Leftrightarrow V = \pi \int\limits_{ - \frac{1}{2}\left( {R - r} \right)}^{\frac{1}{2}\left( {R - r} \right)} {\left[ {{{\left( {\frac{1}{2}\left( {R + r} \right) + \sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}} } \right)}^2} - {{\left( {\frac{1}{2}\left( {R + r} \right) - \sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}} } \right)}^2}} \right]} dx\\ \Leftrightarrow V = \pi \int\limits_{ - \frac{1}{2}\left( {R - r} \right)}^{\frac{1}{2}\left( {R - r} \right)} {\left[ {\left( {\frac{1}{4}{{\left( {R + r} \right)}^2} + \left( {R + r} \right)\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}}  + \frac{1}{4}{{\left( {R - r} \right)}^2} - {x^2}} \right)} \right.} \\ - \left. {\left( {\frac{1}{4}{{\left( {R + r} \right)}^2} - \left( {R + r} \right)\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}}  + \frac{1}{4}{{\left( {R - r} \right)}^2} - {x^2}} \right)} \right]dx\\ \Leftrightarrow V = \pi \int\limits_{ - \frac{1}{2}\left( {R - r} \right)}^{\frac{1}{2}\left( {R - r} \right)} {\left[ {\frac{1}{4}{{\left( {R + r} \right)}^2} + \left( {R + r} \right)\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}}  + \frac{1}{4}{{\left( {R - r} \right)}^2} - {x^2}} \right.} \\\left. { - \frac{1}{4}{{\left( {R + r} \right)}^2} + \left( {R + r} \right)\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}}  - \frac{1}{4}{{\left( {R - r} \right)}^2} + {x^2}} \right]dx\\ \Leftrightarrow V = \pi \int\limits_{ - \frac{1}{2}\left( {R - r} \right)}^{\frac{1}{2}\left( {R - r} \right)} {2\left( {R + r} \right)} \sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}} dx\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\int\limits_{ - \frac{1}{2}\left( {R - r} \right)}^{\frac{1}{2}\left( {R - r} \right)} {\left( {\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}} } \right)} dx\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\left[ {\frac{1}{2}x\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {x^2}}  + \frac{1}{2}.{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}{{\sin }^{ - 1}}\left( {\frac{x}{{\frac{1}{2}\left( {R - r} \right)}}} \right)} \right]_{ - \frac{1}{2}\left( {R - r} \right)}^{\frac{1}{2}\left( {R - r} \right)}\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\left[ {\left( {\frac{1}{2}.\frac{1}{2}\left( {R - r} \right)\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}}  + \frac{1}{2}.{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}{{\sin }^{ - 1}}\left( {\frac{{\frac{1}{2}\left( {R - r} \right)}}{{\frac{1}{2}\left( {R - r} \right)}}} \right)} \right)} \right.\\ - \left. {\left( {\frac{1}{2}. - \frac{1}{2}\left( {R - r} \right)\sqrt {{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2} - {{\left( { - \frac{1}{2}\left( {R - r} \right)} \right)}^2}}  + \frac{1}{2}.{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}{{\sin }^{ - 1}}\left( {\frac{{ - \frac{1}{2}\left( {R - r} \right)}}{{\frac{1}{2}\left( {R - r} \right)}}} \right)} \right)} \right]\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\left[ {\left( {0 + \frac{1}{2}{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}{{\sin }^{ - 1}}\left( 1 \right)} \right) - \left( {0 + \frac{1}{2}{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}{{\sin }^{ - 1}}\left( { - 1} \right)} \right)} \right]\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\left[ {\left( {\frac{1}{2}{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}.\frac{\pi }{2}} \right) - \left( {\frac{1}{2}{{\left( {\frac{1}{2}\left( {R - r} \right)} \right)}^2}. - \frac{\pi }{2}} \right)} \right]\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\left[ {\frac{\pi }{4}.\frac{1}{4}{{\left( {R - r} \right)}^2} + \frac{\pi }{4}.\frac{1}{4}{{\left( {R - r} \right)}^2}} \right]\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\left[ {\frac{\pi }{{16}}{{\left( {R - r} \right)}^2} + \frac{\pi }{{16}}{{\left( {R - r} \right)}^2}} \right]\\ \Leftrightarrow V = 2\pi \left( {R + r} \right)\left[ {\frac{\pi }{8}{{\left( {R - r} \right)}^2}} \right]\\ \Leftrightarrow V = \frac{1}{4}{\pi ^2}\left( {R + r} \right){\left( {R - r} \right)^2}\end{array}$

Akhirnya rumus yang kita cari ketemu juga.

KESIMPULAN

Volume bangun Torus yang memiliki jari-jari luar (R) dan jari-Jari dalam (r) adalah

\[V = \frac{1}{4}{\pi ^2}\left( {R + r} \right){\left( {R - r} \right)^2}\]

Demikian pembahasannya dan terima kasih.