{“appState”:{“pageLoadApiCallsStatus”:true},”articleState”:{“article”:{“headers”:{“creationTime”:”2016-03-26T21:05:45+00:00″,”modifiedTime”:”2023-08-09T16:18:34+00:00″,”timestamp”:”2023-08-09T18:01:03+00:00″},”data”:{“breadcrumbs”:[{“name”:”Academics & The Arts”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33662″},”slug”:”academics-the-arts”,”categoryId”:33662},{“name”:”Math”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33720″},”slug”:”math”,”categoryId”:33720},{“name”:”Calculus”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33723″},”slug”:”calculus”,”categoryId”:33723}],”title”:”How to Find the Area of a Surface of Revolution”,”strippedTitle”:”how to find the area of a surface of revolution”,”slug”:”how-to-find-the-area-of-a-surface-of-revolution”,”canonicalUrl”:””,”seo”:{“metaDescription”:”The calculus instructions in this article and video show you how to find the area of a surface of revolution.”,”noIndex”:0,”noFollow”:0},”content”:”A surface of revolution is a three-dimensional surface with circular cross sections, like a vase or a bell, or a wine bottle. This article, and the video, show you how to find its area.\r\n<div class=\”x2 x2-top\”>\r\n\r\n<div class=\”video-player-organism\”></div>\r\n\r\n</div>\r\nFor these problems, you divide the surface into narrow circular bands, figure the surface area of a representative band, and then just add up the areas of all the bands to get the total surface area. The following figure shows such a shape with a representative band.\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219842.image0.jpg\” alt=\”image0.jpg\” width=\”400\” height=\”225\” />\r\n\r\nWhat’s the surface area of a representative band? Well, if you cut the band and unroll it, you get sort of a long, narrow rectangle whose area, of course, is <i>length</i> times <i>width</i>.\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219843.image1.png\” alt=\”image1.png\” width=\”503\” height=\”221\” />\r\n\r\n<b>Surface of Revolution:</b><b> </b>A surface generated by revolving a function, <i>y</i> = <i>f</i> (<i>x</i>), about an axis has a surface area — between <i>a</i> and <i>b</i> — given by the following integral:\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219844.image2.png\” alt=\”image2.png\” width=\”268\” height=\”59\” />\r\n\r\nBy the way, in the above explanation, you might be wondering why the width of the rectangular band is\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219845.image3.png\” alt=\”image3.png\” width=\”207\” height=\”56\” />\r\n\r\nIt’s because the little band width is slanted instead of horizontal (in which case it would be just <i>dx</i>). The fact that it’s slanted makes it work like the hypotenuse of a little right triangle. The fancy-looking expression for the width of the band comes from working out the length of this hypotenuse with the Pythagorean Theorem. That should make you feel a lot better!\r\n\r\nIf the axis of revolution is the <i>x</i>-axis, <i>r</i> will equal <i>f</i> (<i>x</i>) — as shown in the above figure. If the axis of revolution is some other line, like <i>y</i> = 5, it’s a bit more complicated — something to look forward to.\r\n\r\nNow try a problem: What’s the surface area — between <i>x</i> = 1 and <i>x</i> = 2 — of the surface generated by revolving\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219846.image4.png\” alt=\”image4.png\” width=\”45\” height=\”27\” />\r\n\r\nabout the <i>x</i>-axis?\r\n<div class=\”imageBlock\” style=\”width: 314px;\”>\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219847.image5.jpg\” alt=\”A surface of revolution — this one’s shaped sort of like the end of a trumpet.\” width=\”314\” height=\”400\” />\r\n<div class=\”imageCaption\”>A surface of revolution — this one’s shaped sort of like the end of a trumpet.</div>\r\n</div>\r\n<ol class=\”level-one\”>\r\n \t<li>\r\n<p class=\”first-para\”>Take the derivative of your function.</p>\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219848.image6.png\” alt=\”image6.png\” width=\”63\” height=\”67\” />\r\n<p class=\”child-para\”>Now you can finish the problem by just plugging everything into the formula, but you should do it step by step to reinforce the idea that whenever you integrate, you write down a representative little bit of something — that’s the integrand — then you add up all the little bits by integrating.</p>\r\n</li>\r\n \t<li>\r\n<p class=\”first-para\”>Figure the surface area of a representative narrow band.</p>\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219849.image7.png\” alt=\”image7.png\” width=\”500\” height=\”128\” /></li>\r\n \t<li>\r\n<p class=\”first-para\”>Add up the areas of all the bands from 1 to 2 by integrating.</p>\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219850.image8.png\” alt=\”image8.png\” width=\”473\” height=\”461\” /></li>\r\n</ol>”,”description”:”A surface of revolution is a three-dimensional surface with circular cross sections, like a vase or a bell, or a wine bottle. This article, and the video, show you how to find its area.\r\n<div class=\”x2 x2-top\”>\r\n\r\n<div class=\”video-player-organism\”></div>\r\n\r\n</div>\r\nFor these problems, you divide the surface into narrow circular bands, figure the surface area of a representative band, and then just add up the areas of all the bands to get the total surface area. The following figure shows such a shape with a representative band.\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219842.image0.jpg\” alt=\”image0.jpg\” width=\”400\” height=\”225\” />\r\n\r\nWhat’s the surface area of a representative band? Well, if you cut the band and unroll it, you get sort of a long, narrow rectangle whose area, of course, is <i>length</i> times <i>width</i>.\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219843.image1.png\” alt=\”image1.png\” width=\”503\” height=\”221\” />\r\n\r\n<b>Surface of Revolution:</b><b> </b>A surface generated by revolving a function, <i>y</i> = <i>f</i> (<i>x</i>), about an axis has a surface area — between <i>a</i> and <i>b</i> — given by the following integral:\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219844.image2.png\” alt=\”image2.png\” width=\”268\” height=\”59\” />\r\n\r\nBy the way, in the above explanation, you might be wondering why the width of the rectangular band is\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219845.image3.png\” alt=\”image3.png\” width=\”207\” height=\”56\” />\r\n\r\nIt’s because the little band width is slanted instead of horizontal (in which case it would be just <i>dx</i>). The fact that it’s slanted makes it work like the hypotenuse of a little right triangle. The fancy-looking expression for the width of the band comes from working out the length of this hypotenuse with the Pythagorean Theorem. That should make you feel a lot better!\r\n\r\nIf the axis of revolution is the <i>x</i>-axis, <i>r</i> will equal <i>f</i> (<i>x</i>) — as shown in the above figure. If the axis of revolution is some other line, like <i>y</i> = 5, it’s a bit more complicated — something to look forward to.\r\n\r\nNow try a problem: What’s the surface area — between <i>x</i> = 1 and <i>x</i> = 2 — of the surface generated by revolving\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219846.image4.png\” alt=\”image4.png\” width=\”45\” height=\”27\” />\r\n\r\nabout the <i>x</i>-axis?\r\n<div class=\”imageBlock\” style=\”width: 314px;\”>\r\n\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219847.image5.jpg\” alt=\”A surface of revolution — this one’s shaped sort of like the end of a trumpet.\” width=\”314\” height=\”400\” />\r\n<div class=\”imageCaption\”>A surface of revolution — this one’s shaped sort of like the end of a trumpet.</div>\r\n</div>\r\n<ol class=\”level-one\”>\r\n \t<li>\r\n<p class=\”first-para\”>Take the derivative of your function.</p>\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219848.image6.png\” alt=\”image6.png\” width=\”63\” height=\”67\” />\r\n<p class=\”child-para\”>Now you can finish the problem by just plugging everything into the formula, but you should do it step by step to reinforce the idea that whenever you integrate, you write down a representative little bit of something — that’s the integrand — then you add up all the little bits by integrating.</p>\r\n</li>\r\n \t<li>\r\n<p class=\”first-para\”>Figure the surface area of a representative narrow band.</p>\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219849.image7.png\” alt=\”image7.png\” width=\”500\” height=\”128\” /></li>\r\n \t<li>\r\n<p class=\”first-para\”>Add up the areas of all the bands from 1 to 2 by integrating.</p>\r\n<img src=\”https://www.dummies.com/wp-content/uploads/219850.image8.png\” alt=\”image8.png\” width=\”473\” height=\”461\” /></li>\r\n</ol>”,”blurb”:””,”authors”:[],”primaryCategoryTaxonomy”:{“categoryId”:33723,”title”:”Calculus”,”slug”:”calculus”,”_links”:{“self”:”https://dummies-api.dummies.com/v2/categories/33723″}},”secondaryCategoryTaxonomy”:{“categoryId”:0,”title”:null,”slug”:null,”_links”:null},”tertiaryCategoryTaxonomy”:{“categoryId”:0,”title”:null,”slug”:null,”_links”:null},”trendingArticles”:null,”inThisArticle”:[],”relatedArticles”:{“fromBook”:[],”fromCategory”:[{“articleId”:256336,”title”:”Solve a Difficult Limit Problem Using the Sandwich Method”,”slug”:”solve-a-difficult-limit-problem-using-the-sandwich-method”,”categoryList”:[“academics-the-arts”,”math”,”calculus”],”_links”:{“self”:”https://dummies-api.dummies.com/v2/articles/256336″}},{“articleId”:255765,”title”:”Solve Limit Problems on a Calculator Using Graphing Mode”,”slug”:”solve-limit-problems-on-a-calculator-using-graphing-mode”,”categoryList”:[“academics-the-arts”,”math”,”calculus”],”_links”:{“self”:”https://dummies-api.dummies.com/v2/articles/255765″}},{“articleId”:255755,”title”:”Solve Limit Problems on a Calculator Using the Arrow-Number”,”slug”:”solve-limit-problems-on-a-calculator-using-the-arrow-number”,”categoryList”:[“academics-the-arts”,”math”,”calculus”],”_links”:{“self”:”https://dummies-api.dummies.com/v2/articles/255755″}},{“articleId”:255261,”title”:”Limit and Continuity Graphs: Practice Questions”,”slug”:”limit-and-continuity-graphs-practice-questions”,”categoryList”:[“academics-the-arts”,”math”,”calculus”],”_links”:{“self”:”https://dummies-api.dummies.com/v2/articles/255261″}},{“articleId”:255255,”title”:”Use the Vertical Line Test to Identify a Function”,”slug”:”use-the-vertical-line-test-to-identify-a-function”,”categoryList”:[“academics-the-arts”,”math”,”calculus”],”_links”:{“self”:”https://dummies-api.dummies.com/v2/articles/255255″}}]},”hasRelatedBookFromSearch”:true,”relatedBook”:{“bookId”:282153,”slug”:”differential-equations-for-dummies”,”isbn”:”9780470178140″,”categoryList”:[“academics-the-arts”,”math”,”calculus”],”amazon”:{“default”:”https://www.amazon.com/gp/product/0470178140/ref=as_li_tl?ie=UTF8&tag=wiley01-20″,”ca”:”https://www.amazon.ca/gp/product/0470178140/ref=as_li_tl?ie=UTF8&tag=wiley01-20″,”indigo_ca”:”http://www.tkqlhce.com/click-9208661-13710633?url=https://www.chapters.indigo.ca/en-ca/books/product/0470178140-item.html&cjsku=978111945484″,”gb”:”https://www.amazon.co.uk/gp/product/0470178140/ref=as_li_tl?ie=UTF8&tag=wiley01-20″,”de”:”https://www.amazon.de/gp/product/0470178140/ref=as_li_tl?ie=UTF8&tag=wiley01-20″},”image”:{“src”:”https://www.dummies.com/covers/9780470178140.jpg”,”width”:250,”height”:350},”title”:”Differential Equations For Dummies”,”testBankPinActivationLink”:””,”bookOutOfPrint”:false,”authorsInfo”:”\n <p><p><b><b data-author-id=\”8967\”>Steven Holzner</b>, PhD,</b> taught physics at Cornell University for more than 10 years.</p>”,”authors”:[{“authorId”:8967,”name”:”Steven Holzner”,”slug”:”steven-holzner”,”description”:” <p><b>Steven Holzner, PhD,</b> taught physics at Cornell University for more than 10 years. “,”hasArticle”:false,”_links”:{“self”:”https://dummies-api.dummies.com/v2/authors/8967″}}],”_links”:{“self”:”https://dummies-api.dummies.com/v2/books/282153″}},”collections”:[],”articleAds”:{“footerAd”:”<div class=\”du-ad-region row\” id=\”article_page_adhesion_ad\”><div class=\”du-ad-unit col-md-12\” data-slot-id=\”article_page_adhesion_ad\” data-refreshed=\”false\” \r\n data-target = \”[{"key":"cat","values":["academics-the-arts","math","calculus"]},{"key":"isbn","values":[null]}]\” id=\”du-slot-64d3d45f0a144\”></div></div>”,”rightAd”:”<div class=\”du-ad-region row\” 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Formulas in this calculus video tutorial reveal how to estimate, measure, and solve for the surface area of a three-dimensional object like a”,”uploadDate”:”2022-07-15T08:18:07.281Z”}},”sponsorship”:{“sponsorshipPage”:false,”backgroundImage”:{“src”:null,”width”:0,”height”:0},”brandingLine”:””,”brandingLink”:””,”brandingLogo”:{“src”:null,”width”:0,”height”:0},”sponsorAd”:””,”sponsorEbookTitle”:””,”sponsorEbookLink”:””,”sponsorEbookImage”:{“src”:null,”width”:0,”height”:0}},”primaryLearningPath”:”Advance”,”lifeExpectancy”:”Five 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