{"id":409,"date":"2019-06-26T23:56:52","date_gmt":"2019-06-26T23:56:52","guid":{"rendered":"http:\/\/funfacts.104.42.120.246.xip.io\/?page_id=409"},"modified":"2020-01-03T22:23:59","modified_gmt":"2020-01-03T22:23:59","slug":"surface-area-of-a-sphere","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/funfacts\/surface-area-of-a-sphere\/","title":{"rendered":"Surface Area of a Sphere"},"content":{"rendered":"\n<p>The area of a disk enclosed by a circle of radius R is Pi*R<sup>2<\/sup>.\u00a0<br>The formula for the circumference of a circle of radius R is 2*Pi*R.\u00a0<br>A simple\u00a0calculus\u00a0check reveals that the latter is the derivative of the former with respect to R.<\/p>\n\n\n\n<p>Similarly, the volume of a ball enclosed by a sphere of radius R is (4\/3)*Pi*R<sup>3<\/sup>.&nbsp;<br>And the formula for the surface area of a sphere of radius R is 4*Pi*R<sup>2<\/sup>.&nbsp;<br>And, you can check that the latter is the derivative of the former with respect to R.<\/p>\n\n\n\n<p>Coincidence, or is there a reason?<\/p>\n\n\n\n<p><strong>Presentation\u00a0Suggestions:<\/strong><br>Let your students tell you those\u00a0geometry\u00a0formulas if they remember them.<\/p>\n\n\n\n<p><strong>The&nbsp;Math&nbsp;Behind&nbsp;the&nbsp;Fact:<\/strong><br>Well, no, it is not a coincidence. For the ball, a small change in radius produces a change in volume of the ball which is equal to the volume of a spherical shell of radius R and thickness (delta R). The spherical shell&#8217;s volume is thus approximately (surface area of the sphere)*(delta R). But the derivative is approximately the change in ball volume divided by (delta R), which is thus just (surface area of the sphere).<\/p>\n\n\n\n<p>So, if I tell you the 4-dimensional &#8220;volume&#8221; of the 4-dimensional ball is (1\/2)*Pi<sup>2<\/sup>*R<sup>4<\/sup>, what is 3-dimensional volume of its boundary?<\/p>\n\n\n\n<p>See also\u00a0Volume of a Ball in N Dimensions.<\/p>\n\n\n\n<p><strong>How to Cite this Page:<\/strong>&nbsp;<br>Su, Francis E., et al. &#8220;Surface Area of a Sphere.&#8221;&nbsp;<em>Math Fun Facts<\/em>. &lt;http:\/\/www.math.hmc.edu\/funfacts&gt;.<\/p>\n\n\n\n<p><strong>Fun Fact suggested by: <\/strong><br>Francis Su <\/p>\n","protected":false},"excerpt":{"rendered":"<p>The area of a disk enclosed by a circle of radius R is Pi*R2.\u00a0The formula for the circumference of a&#46;&#46;&#46;<\/p>\n","protected":false},"author":7,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[20,7,8,4,44],"class_list":["post-409","page","type-page","status-publish","hentry","tag-analysis","tag-calculus","tag-geometry","tag-medium","tag-sphere"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/409","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/comments?post=409"}],"version-history":[{"count":3,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/409\/revisions"}],"predecessor-version":[{"id":1670,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/409\/revisions\/1670"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/media?parent=409"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/tags?post=409"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}