{"id":257,"date":"2002-09-11T09:18:00","date_gmt":"2002-09-11T14:18:00","guid":{"rendered":"http:\/\/cheerfulchaotic.crazycrew.org\/?p=257"},"modified":"2002-09-11T09:18:00","modified_gmt":"2002-09-11T14:18:00","slug":"257","status":"publish","type":"post","link":"https:\/\/cheerfulchaotic.crazycrew.org\/?p=257","title":{"rendered":""},"content":{"rendered":"<p>WTF happened to my brain?<\/p>\n<p>So, last night, <lj user=eomsam> was over here working on his math homework.  And the only problem I was called in on was actually pretty simple, yet I totally blew it.  Say you have two circles, of radii R and r, such that R > r.  The smaller circle is partially within the larger circle, such that the line segment formed by the points of intersection between the two circles is the diameter of the smaller circle.  Find the area of the smaller circle that is outside the larger circle.  <\/p>\n<p>And I tried to integrate.  Clearly I&#8217;m a moron.  That is so unneccesary.  <\/p>\n<p>Consider the triangle formed by the center of the larger circle and the points of intersection of the smaller circle, it has two sides of length R, and one side of length 2r.  So we know that the angle formed by the two R sides is 2 x arcsin(r\/R).  If we multiply the fraction of that angle out of the full circle (360, 2*pi, whatever, take your pick) by the area of the large circle (pi*R^2), you get the area of the wedge, subtract from that the area of the 2r,R,R triangle, (r*sqrt(R^2-r^2)) and you have the area of the circular end of the wedge.  Subtract that from half the area of the small circle (pi*r^2\/2) and you have the area of the lune.  Duh.<\/p>\n<p>Spelled out:  pi*r^2\/2 &#8211; (2*arcsin(r\/R)*pi*R^2)\/(2*pi) + r*sqrt(R^2-r^2)<br \/>\nor<br \/>\npi*r^2\/2 &#8211; arcsin(r\/R)*R^2 + r*sqrt(R^2-r^2)<\/p>\n<p>Stupid math problem.  Grrr.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>WTF happened to my brain? So, last night, was over here working on his math homework. And the only problem I was called in on was actually pretty simple, yet I totally blew it. Say you have two circles, of radii R and r, such that R > r. The smaller circle is partially within &hellip; <a href=\"https:\/\/cheerfulchaotic.crazycrew.org\/?p=257\" class=\"more-link\">Continue reading<span class=\"screen-reader-text\"> &#8220;&#8221;<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-257","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=\/wp\/v2\/posts\/257","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=257"}],"version-history":[{"count":0,"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=\/wp\/v2\/posts\/257\/revisions"}],"wp:attachment":[{"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=257"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=257"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cheerfulchaotic.crazycrew.org\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}