{"id":1202,"date":"2019-05-18T19:13:49","date_gmt":"2019-05-18T17:13:49","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=1202"},"modified":"2022-06-13T16:18:09","modified_gmt":"2022-06-13T14:18:09","slug":"pourquoi-laire-dun-disque-est-egale-a-pi-r2","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2019\/05\/18\/pourquoi-laire-dun-disque-est-egale-a-pi-r2\/","title":{"rendered":"Pourquoi l’aire d’un disque est \u00e9gale \u00e0 \\(\\pi r^2\\) ?"},"content":{"rendered":"\n

Consid\u00e9rons un disque de rayon r<\/em>. Nous allons rapport\u00e9 le plan \u00e0 un rep\u00e8re orthonorm\u00e9 d’origine O<\/em>, et nous allons centrer notre disque en O<\/em>.<\/p>\n\n\n

\n
\"\"<\/a>
Disque de centre O et de rayon r<\/figcaption><\/figure>\n<\/div>\n\n\n\n\n\n\n

Afin de d\u00e9terminer l’aire du disque, consid\u00e9rons uniquement son enveloppe : le cercle de centre O<\/em> et de rayon r<\/em> :<\/p>\n\n\n

\n
\"\"<\/a>
Cercle de centre O et de rayon r<\/figcaption><\/figure>\n<\/div>\n\n\n

Maintenant, consid\u00e9rons un point M<\/em> sur ce cercle d’abscisse x<\/em> > 0 et d’ordonn\u00e9e y<\/em> > 0. Alors, d’apr\u00e8s le th\u00e9or\u00e8me de Pythagore,$$x^2+y^2=r^2.$$<\/p>\n\n\n

\n
\"\"<\/a>
Le point M est sur le cercle<\/figcaption><\/figure>\n<\/div>\n\n\n

Ainsi,$$y=\\sqrt{r^2-x^2}.$$On peut donc consid\u00e9rer le quart de cercle sup\u00e9rieur droit comme la repr\u00e9sentation graphique de la fonction $f$ d\u00e9finie par:$$f(x)=\\sqrt{r^2-x^2}$$sur [0;r<\/em>].<\/p>\n\n\n

\n
\"\"<\/a><\/figure>\n<\/div>\n\n\n

Pour conna\u00eetre l’aire du domaine d\u00e9limit\u00e9 par l’axe des abscisses et la courbe repr\u00e9sentative de f<\/em> sur [0;r<\/em>], il existe un outil math\u00e9matique : l’int\u00e9gration. L’aire cherch\u00e9e est:$$I=\\int_0^rf(x)\\text{d}x=\\int_0^r\\sqrt{r^2-x^2}\\text{d}x.$$<\/p>\n\n\n\n

Pour calculer une int\u00e9grale, il suffit de conna\u00eetre une primitive de la fonction, mais en terminale, nous ne connaissons pas de primitive \u00e0 la fonction f<\/em>. On va donc utiliser une m\u00e9thode pour arriver \u00e0 nos fins (qui n’est pas au programme de Terminale, rassurez-vous).<\/p>\n\n\n\n

Nous allons d’abord \u00e9crire f<\/em>(x<\/em>) autrement:$$\\sqrt{r^2-x^2}=r\\sqrt{1-\\left(\\frac{x}{r}\\right)^2}.$$On peut ainsi \u00e9crire:$$I=r\\int_0^r\\sqrt{1-
\\left(\\frac{x}{r}\\right)^2}\\text{d}x.$$<\/p>\n\n\n\n

Consid\u00e9rons alors une variable u<\/em> telle que:$$u=\\frac{x}{r}.$$Si x<\/em> varie de 0 \u00e0 r<\/em> alors u<\/em> varie de 0 \u00e0 1. De plus, si on consid\u00e8re u<\/em> comme une fonction de x<\/em> alors sa d\u00e9riv\u00e9e par rapport \u00e0 x<\/em> est:$$u'(x)=\\frac{\\text{d}u}{\\text{d}x}=\\frac{1}{r}.$$On peut alors \u00e9crire:$$\\text{d}x=r\\text {d}u.$$Ainsi, si nous voulons exprimer l’int\u00e9grale I<\/em> en fonction de u<\/em>, on \u00e9crit:$$I=r\\int_0^1\\sqrt{1-u^2}r\\text{d}u=r^2\\int_0^1\\sqrt{1-u^2}\\text{d}u .$$<\/p>\n\n\n\n

Maintenant, la variable d’int\u00e9gration est u<\/em>, et varie de 0 \u00e0 1. On peut donc dire que c’est un sinus (par exemple). Posons alors:$$u=\\sin(t).$$Alors,$$\\text{d}u=\\cos(t)\\text{d}t.$$De plus, si u<\/em> varie de 0 \u00e0 1 alors t<\/em> varie de 0 \u00e0 \\(\\frac{\\pi}{2}\\), d’o\u00f9:$$I=r^2\\int_0^{\\frac{\\pi}{2}}\\sqrt{1-\\sin^2(t)}\\cos(t)\\text{d}t=r^2\\int_0^{\\frac{\\pi}{2}}\\cos^2(t)\\text{d}t.$$En effet, \\(\\cos^2(t)+\\sin^2(t)=1\\) donc \\(1-\\sin^2(t)=\\cos^2(t)\\); et comme t<\/em> varie entre 0 et \\(\\frac{\\pi}{2}\\), son cosinus est positif donc \\(\\sqrt{\\cos^2(t)})\\cos(t)\\) dans notre int\u00e9grale.<\/p>\n\n\n\n

Il ne reste plus qu’\u00e0 trouver une primitive de \\(\\cos^2(t)\\). Pour cela, il faut se souvenir que:$$\\cos(2t)=2\\cos^2(t)-1$$et donc que:$$\\cos^2(t)=\\frac{1}{2}(\\cos(2t)+1).$$Ainsi,$$\\begin{align}I&=r^2\\int_0^{\\frac{\\pi}{2}}\\frac{1}{2}(\\cos(2t)+1)\\text{d}t\\\\&=\\frac{r^2}{2}\\int_0^{\\frac{\\pi}{2}}(\\cos(2t)+1)\\text{d}t\\\\&=\\frac{r^2}{2}\\left[ \\frac{1}{2}\\sin(2t)+t\\right]_0^{\\pi\/2}\\\\&=\\frac{r^2}{2}\\times\\frac{\\pi}{2}\\\\&=\\frac{\\pi r^2}{4}. \\end{align}$$<\/p>\n\n\n\n

L’aire du disque \u00e9tant \u00e9gale au quadruple de l’aire trouv\u00e9e, on obtient finalement que l’aire du disque est \u00e9gale \u00e0 :$$\\mathcal{A}_{\\mathcal{D}}=4I=\\pi r^2.$$<\/p>\n\n\n\n

Une pr\u00e9cision sur l’outil d’int\u00e9gration<\/h2>\n\n\n
\n
\"\"<\/a><\/figure>\n<\/div>\n\n\n

Pour un r\u00e9el x<\/em> quelconque de l’intervalle [0;r<\/em>], le segment trac\u00e9 a une longueur \u00e9gale \u00e0 f<\/em>(x<\/em>).<\/p>\n\n\n\n

Trouver l’aire du quart de cercle revient \u00e0 \u00ab\u00a0additionner\u00a0\u00bb la longueur de tous les segments obtenus en faisant varier x<\/em> de 0 \u00e0 r<\/em>. Mais x<\/em> est un nombre r\u00e9el donc il existe une infinit\u00e9 de valeurs. On ne peut donc pas additionner \u00ab\u00a0une \u00e0 une\u00a0\u00bb toutes ces longueurs. On dit que la somme n’est pas discr\u00e8te<\/em> (une somme discr\u00e8te est une somme o\u00f9 l’on peut compter un \u00e0 un tous ses termes, m\u00eame s’il y en a une infinit\u00e9, comme 1+2+3+…). La somme est qualifi\u00e9e de continue<\/em>. Et donc l’int\u00e9grale repr\u00e9sente une somme continue.<\/p>\n\n\n

\n
\"\"<\/a>
Repr\u00e9sentation sch\u00e9matique d’une somme continue des longueurs d segments sur [0 ; r]<\/figcaption><\/figure>\n<\/div>\n\n\n

Sur le sch\u00e9ma ci-dessus, il faut imaginer que l’on rapproche de plus en plus les segments jusqu’\u00e0 ce qu’ils soient tous \u00ab\u00a0coll\u00e9s\u00a0\u00bb. Ils couvrent alors toute la surface. Ainsi, la somme de leurs longueurs sera \u00e9gale \u00e0 l’aire.<\/p>\n\n\n\n

Avec cela, nous pouvons continuer et nous demander pourquoi le volume d’une boule est \u00e9gal \u00e0 \\(\\frac{4}{3}\\pi r^3\\) : c’est l’objet de l’article suivant<\/a>.<\/p>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

Consid\u00e9rons un disque de rayon r. Nous allons rapport\u00e9 le plan \u00e0 un rep\u00e8re orthonorm\u00e9 d’origine O, et nous allons centrer notre disque en O.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[54,104],"class_list":["post-1202","post","type-post","status-publish","format-standard","hentry","category-mathematiques","tag-aire","tag-integrales"],"yoast_head":"\nPourquoi l'aire d'un disque est \u00e9gale \u00e0 \\(\\pi r^2\\) ? - Mathweb.fr<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/2019\/05\/18\/pourquoi-laire-dun-disque-est-egale-a-pi-r2\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Pourquoi l'aire d'un disque est \u00e9gale \u00e0 \\(\\pi r^2\\) ? - Mathweb.fr\" \/>\n<meta property=\"og:description\" content=\"Consid\u00e9rons un disque de rayon r. 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