{"id":235384,"date":"2025-06-15T10:33:11","date_gmt":"2025-06-15T10:33:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=235384"},"modified":"2025-06-15T10:33:13","modified_gmt":"2025-06-15T10:33:13","slug":"general-slicing-method-use-the-general-slicing-method-to-find-the-volume-of-the-following-solids","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/15\/general-slicing-method-use-the-general-slicing-method-to-find-the-volume-of-the-following-solids\/","title":{"rendered":"General slicing method Use the general slicing method to find the volume of the following solids"},"content":{"rendered":"\n<p>General slicing method Use the general slicing method to find the volume of the following solids. 7. The solid whose base is the region bounded by the curves<br>and<br>and whose cross sections through the solid perpendicular to the x-axis are squares<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-521.png\" alt=\"\" class=\"wp-image-235385\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer:<\/strong>&nbsp;The volume of the solid is&nbsp;<strong>64\/15<\/strong>&nbsp;cubic units.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>This problem requires finding the volume of a solid using the general slicing method. The process involves integrating the area of a cross-section across the length of the solid&#8217;s base.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the Base and Integration Limits:<\/strong><br>The base of the solid is the region bounded by the curves\u00a0y = x\u00b2\u00a0(a parabola opening upwards) and\u00a0y = 2 &#8211; x\u00b2\u00a0(a parabola opening downwards). To find the extent of this region along the x-axis, the intersection points of the curves must be found. This is done by setting the two equations equal to each other:<br>x\u00b2 = 2 &#8211; x\u00b2<br>2x\u00b2 = 2<br>x\u00b2 = 1<br>x = \u00b11<br>These intersection points,\u00a0x = -1\u00a0and\u00a0x = 1, will be the limits of integration.<\/li>\n\n\n\n<li><strong>Define the Cross-Sectional Area:<\/strong><br>The problem specifies that the cross-sections are perpendicular to the x-axis and are squares. The side length,\u00a0s, of each square at a specific x-value is the vertical distance between the two bounding curves. Within the interval [-1, 1], the curve\u00a0y = 2 &#8211; x\u00b2\u00a0is the upper boundary and\u00a0y = x\u00b2\u00a0is the lower boundary.<br>The side length\u00a0s(x)\u00a0is therefore:<br>s(x) = (top curve) &#8211; (bottom curve) = (2 &#8211; x\u00b2) &#8211; (x\u00b2) = 2 &#8211; 2x\u00b2<br>The area of a square is\u00a0side\u00b2, so the area of a single cross-sectional slice,\u00a0A(x), is:<br>A(x) = [s(x)]\u00b2 = (2 &#8211; 2x\u00b2)\u00b2 = 4 &#8211; 8x\u00b2 + 4x\u2074<\/li>\n\n\n\n<li><strong>Set Up and Evaluate the Volume Integral:<\/strong><br>The volume\u00a0V\u00a0of the solid is the integral of the cross-sectional area\u00a0A(x)\u00a0from the lower limit\u00a0x = -1\u00a0to the upper limit\u00a0x = 1.<br>V = \u222b[from -1 to 1] A(x) dx = \u222b[from -1 to 1] (4 &#8211; 8x\u00b2 + 4x\u2074) dx<br>To evaluate this integral, find the antiderivative of the function:<br>\u222b(4 &#8211; 8x\u00b2 + 4x\u2074) dx = 4x &#8211; (8\/3)x\u00b3 + (4\/5)x\u2075<br>Now, apply the Fundamental Theorem of Calculus using the limits of integration:<br>V = [4x &#8211; (8\/3)x\u00b3 + (4\/5)x\u2075] from -1 to 1<br>V = [4(1) &#8211; (8\/3)(1)\u00b3 + (4\/5)(1)\u2075] &#8211; [4(-1) &#8211; (8\/3)(-1)\u00b3 + (4\/5)(-1)\u2075]<br>V = [4 &#8211; 8\/3 + 4\/5] &#8211; [-4 + 8\/3 &#8211; 4\/5]<br>V = 8 &#8211; 16\/3 + 8\/5<br>Finding a common denominator (15):<br>V = 120\/15 &#8211; 80\/15 + 24\/15 = (120 &#8211; 80 + 24) \/ 15 = 64\/15<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-602.jpeg\" alt=\"\" class=\"wp-image-235386\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>General slicing method Use the general slicing method to find the volume of the following solids. 7. The solid whose base is the region bounded by the curvesandand whose cross sections through the solid perpendicular to the x-axis are squares The Correct Answer and Explanation is: Correct Answer:&nbsp;The volume of the solid is&nbsp;64\/15&nbsp;cubic units. Explanation [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-235384","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/235384","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=235384"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/235384\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=235384"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=235384"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=235384"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}