{"id":3789,"date":"2020-10-10T16:48:22","date_gmt":"2020-10-10T14:48:22","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=3789"},"modified":"2024-08-28T08:06:08","modified_gmt":"2024-08-28T06:06:08","slug":"pourquoi-le-volume-dune-boule-est-egal-a-frac43pi-r3-explications-avec-les-integrales","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/10\/pourquoi-le-volume-dune-boule-est-egal-a-frac43pi-r3-explications-avec-les-integrales\/","title":{"rendered":"Pourquoi le volume d’une boule est \u00e9gal \u00e0 \\(\\frac{4}{3}\\pi r^3\\) ? Explications avec les int\u00e9grales"},"content":{"rendered":"\n
Volume d’une boule avec une int\u00e9grale. Ceci est une boule:<\/p>\n\n\n
\n<\/figure>\n<\/div>\n\n\n
Si l’on consid\u00e8re que son rayon est \u00e9gal \u00e0 R<\/em> alors son volume est \\(\\frac{4}{3}\\pi R^3\\)… mais pourquoi ?<\/p>\n\n\n\n\n\n\n\n
Pla\u00e7ons-nous dans un rep\u00e8re orthonorm\u00e9 de l’espace et pla\u00e7ons-y notre boule de sorte que son centre co\u00efncide avec l’origine du rep\u00e8re :<\/p>\n\n\n
\n<\/figure>\n<\/div>\n\n\n
Nous avons aussi introduit un point A de coordonn\u00e9es (0;0;z<\/em>), o\u00f9 z<\/em> varie de –R<\/em> \u00e0 +R<\/em>. Nous avons ensuite consid\u00e9r\u00e9 le disque de centre A et de rayon r<\/em>(z<\/em>), section de la boule et du plan passant par A et parall\u00e8le au plan (x<\/em>Oy<\/em>).<\/p>\n\n\n\n
Le volume de la boule n’est autre que la somme des volumes des cylindres de base \\(\\pi r(z)^2\\) d’\u00e9paisseur infinit\u00e9simale dz, pour z<\/em> variant de –R<\/em> \u00e0 +R<\/em>, somme infinit\u00e9simale donc que l’on peut prendre comme une int\u00e9grale:$$\\mathcal{V}=\\int_{-R}^{+R}\\pi r(z)^2\\text{d}z.$$<\/p>\n\n\n
\n<\/figure>\n<\/div>\n\n\n
Il ne reste plus qu’\u00e0 trouver l’expression de r<\/em>(z<\/em>)… et ce n’est pas trop compliqu\u00e9 car d’apr\u00e8s le th\u00e9or\u00e8me de Pythagore, dans le triangle AOB:$$OB^2=OA^2+AB^2$$soit:$$R^2=z^2+\\big[r(z)\\big]^2$$d’o\u00f9:$$r(z)^2=R^2-z^2.$$<\/p>\n\n\n\n
\u00c7a, c’est fait ! Il faut maintenant se pencher sur le calcul de l’int\u00e9grale:$$\\begin{align}\\mathcal{V}&=\\pi\\int_{-R}^{+R}(R^2-z^2)\\text{d}z\\\\&=\\pi\\left[R^2z – \\frac{1}{3}z^3\\right]_{-R}^{+R}\\\\&=\\pi\\left[\\left(R^3-\\frac{1}{3}R^3\\right) – \\left(-R^3+\\frac{1}{3}R^3\\right)\\right]\\\\&= \\frac{4}{3}\\pi R^3.\\end{align}$$<\/p>\n\n\n\n
On obtient alors la formule connue des coll\u00e9giens ! Myst\u00e8re r\u00e9solu !<\/p>\n","protected":false},"excerpt":{"rendered":"
Volume d’une boule avec une int\u00e9grale. Ceci est une boule: Si l’on consid\u00e8re que son rayon est \u00e9gal \u00e0 R alors son volume est \\(\\frac{4}{3}\\pi R^3\\)… mais pourquoi ?<\/p>\n","protected":false},"author":1,"featured_media":3791,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,6],"tags":[243,104,106,105],"class_list":["post-3789","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-enseignement","category-mathematiques","tag-formule","tag-integrales","tag-sphere","tag-volume"],"yoast_head":"\n
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