{"id":233487,"date":"2025-06-13T07:01:06","date_gmt":"2025-06-13T07:01:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233487"},"modified":"2025-06-13T07:01:09","modified_gmt":"2025-06-13T07:01:09","slug":"find-the-surface-area-generated-by-rotating-the-first-quadrant-portion-of-the-curve-xtextsuperscript216-8y-about-the-y-axis","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/find-the-surface-area-generated-by-rotating-the-first-quadrant-portion-of-the-curve-xtextsuperscript216-8y-about-the-y-axis\/","title":{"rendered":"Find the surface area generated by rotating the first quadrant portion of the curve x\\textsuperscript{2}=16-8y about the y-axis"},"content":{"rendered":"\n

Find the surface area generated by rotating the first quadrant portion of the curve x\\textsuperscript{2}=16-8y about the y-axis. ? 64.25 ? 66.38 ? 61.27 ? 58.41<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the surface area<\/strong> generated by rotating the first-quadrant portion of the curvex2=16\u22128yx^2 = 16 – 8yx2=16\u22128y<\/p>\n\n\n\n

about the y-axis<\/strong>, we use the formula for surface area of revolution:A=2\u03c0\u222by1y2x1+(dxdy)2\u2009dyA = 2\\pi \\int_{y_1}^{y_2} x \\sqrt{1 + \\left(\\frac{dx}{dy}\\right)^2} \\, dyA=2\u03c0\u222by1\u200by2\u200b\u200bx1+(dydx\u200b)2\u200bdy<\/p>\n\n\n\n

\n\n\n\n

Step 1: Solve for xxx in terms of yyy<\/strong><\/h3>\n\n\n\n

Given:x2=16\u22128y\u21d2x=16\u22128yx^2 = 16 – 8y \\Rightarrow x = \\sqrt{16 – 8y}x2=16\u22128y\u21d2x=16\u22128y\u200b<\/p>\n\n\n\n

(Since we’re in the first quadrant, x\u22650x \\geq 0x\u22650)<\/p>\n\n\n\n

\n\n\n\n

Step 2: Find bounds for yyy<\/strong><\/h3>\n\n\n\n

The curve intersects the x-axis when y=0y = 0y=0:x2=16\u21d2x=4x^2 = 16 \\Rightarrow x = 4×2=16\u21d2x=4<\/p>\n\n\n\n

The curve intersects the y-axis when x=0x = 0x=0:0=16\u22128y\u21d2y=20 = \\sqrt{16 – 8y} \\Rightarrow y = 20=16\u22128y\u200b\u21d2y=2<\/p>\n\n\n\n

So we integrate from y=0y = 0y=0 to y=2y = 2y=2.<\/p>\n\n\n\n

\n\n\n\n

Step 3: Find dxdy\\frac{dx}{dy}dydx\u200b<\/strong><\/h3>\n\n\n\n

x=(16\u22128y)1\/2\u21d2dxdy=12(16\u22128y)\u22121\/2\u22c5(\u22128)=\u2212416\u22128yx = (16 – 8y)^{1\/2} \\Rightarrow \\frac{dx}{dy} = \\frac{1}{2}(16 – 8y)^{-1\/2} \\cdot (-8) = \\frac{-4}{\\sqrt{16 – 8y}}x=(16\u22128y)1\/2\u21d2dydx\u200b=21\u200b(16\u22128y)\u22121\/2\u22c5(\u22128)=16\u22128y\u200b\u22124\u200b<\/p>\n\n\n\n

\n\n\n\n

Step 4: Plug into surface area formula<\/strong><\/h3>\n\n\n\n

A=2\u03c0\u222b0216\u22128y\u22c51+(416\u22128y)2\u2009dyA = 2\\pi \\int_0^2 \\sqrt{16 – 8y} \\cdot \\sqrt{1 + \\left(\\frac{4}{\\sqrt{16 – 8y}}\\right)^2} \\, dyA=2\u03c0\u222b02\u200b16\u22128y\u200b\u22c51+(16\u22128y\u200b4\u200b)2\u200bdy<\/p>\n\n\n\n

Simplify the expression inside the integral:(416\u22128y)2=1616\u22128y\\left(\\frac{4}{\\sqrt{16 – 8y}}\\right)^2 = \\frac{16}{16 – 8y}(16\u22128y\u200b4\u200b)2=16\u22128y16\u200b<\/p>\n\n\n\n

So:A=2\u03c0\u222b0216\u22128y\u22c51+1616\u22128y\u2009dy=2\u03c0\u222b0216\u22128y\u22c516\u22128y+1616\u22128y\u2009dyA = 2\\pi \\int_0^2 \\sqrt{16 – 8y} \\cdot \\sqrt{1 + \\frac{16}{16 – 8y}} \\, dy = 2\\pi \\int_0^2 \\sqrt{16 – 8y} \\cdot \\sqrt{\\frac{16 – 8y + 16}{16 – 8y}} \\, dyA=2\u03c0\u222b02\u200b16\u22128y\u200b\u22c51+16\u22128y16\u200b\u200bdy=2\u03c0\u222b02\u200b16\u22128y\u200b\u22c516\u22128y16\u22128y+16\u200b\u200bdy=2\u03c0\u222b0216\u22128y\u22c532\u22128y16\u22128y\u2009dy= 2\\pi \\int_0^2 \\sqrt{16 – 8y} \\cdot \\sqrt{\\frac{32 – 8y}{16 – 8y}} \\, dy=2\u03c0\u222b02\u200b16\u22128y\u200b\u22c516\u22128y32\u22128y\u200b\u200bdy=2\u03c0\u222b0232\u22128y\u2009dy= 2\\pi \\int_0^2 \\sqrt{32 – 8y} \\, dy=2\u03c0\u222b02\u200b32\u22128y\u200bdy<\/p>\n\n\n\n

Let\u2019s substitute:
Let u=32\u22128y\u21d2du=\u22128dy\u21d2dy=\u221218duu = 32 – 8y \\Rightarrow du = -8dy \\Rightarrow dy = -\\frac{1}{8}duu=32\u22128y\u21d2du=\u22128dy\u21d2dy=\u221281\u200bdu
When y=0y = 0y=0, u=32u = 32u=32; when y=2y = 2y=2, u=16u = 16u=16A=2\u03c0\u222b3216u\u22c5(\u221218)du=\u03c04\u222b1632u1\/2duA = 2\\pi \\int_{32}^{16} \\sqrt{u} \\cdot \\left(-\\frac{1}{8}\\right) du = \\frac{\\pi}{4} \\int_{16}^{32} u^{1\/2} duA=2\u03c0\u222b3216\u200bu\u200b\u22c5(\u221281\u200b)du=4\u03c0\u200b\u222b1632\u200bu1\/2du<\/p>\n\n\n\n

Now integrate:\u03c04\u22c5[23u3\/2]1632=\u03c04\u22c523(323\/2\u2212163\/2)\\frac{\\pi}{4} \\cdot \\left[\\frac{2}{3} u^{3\/2} \\right]_{16}^{32} = \\frac{\\pi}{4} \\cdot \\frac{2}{3} \\left(32^{3\/2} – 16^{3\/2} \\right)4\u03c0\u200b\u22c5[32\u200bu3\/2]1632\u200b=4\u03c0\u200b\u22c532\u200b(323\/2\u2212163\/2)=\u03c06(323\u2212163)=\u03c06((42)3\u221243)= \\frac{\\pi}{6} \\left( \\sqrt{32}^3 – \\sqrt{16}^3 \\right) = \\frac{\\pi}{6} \\left( (4\\sqrt{2})^3 – 4^3 \\right)=6\u03c0\u200b(32\u200b3\u221216\u200b3)=6\u03c0\u200b((42\u200b)3\u221243)=\u03c06(642\u221264)=64\u03c06(2\u22121)= \\frac{\\pi}{6} \\left( 64\\sqrt{2} – 64 \\right) = \\frac{64\\pi}{6} (\\sqrt{2} – 1)=6\u03c0\u200b(642\u200b\u221264)=664\u03c0\u200b(2\u200b\u22121)=32\u03c03(2\u22121)= \\frac{32\\pi}{3} (\\sqrt{2} – 1)=332\u03c0\u200b(2\u200b\u22121)<\/p>\n\n\n\n

Now approximate:2\u22481.4142\u21d2(2\u22121)\u22480.4142\\sqrt{2} \\approx 1.4142 \\Rightarrow (\\sqrt{2} – 1) \\approx 0.41422\u200b\u22481.4142\u21d2(2\u200b\u22121)\u22480.4142A\u224832\u03c03\u22c50.4142\u2248100.533\u224833.51A \\approx \\frac{32\\pi}{3} \\cdot 0.4142 \\approx \\frac{100.53}{3} \\approx 33.51A\u2248332\u03c0\u200b\u22c50.4142\u22483100.53\u200b\u224833.51A\u224833.51\u22c5\u03c0\u224833.51\u22c53.1416\u2248105.19A \\approx 33.51 \\cdot \\pi \\approx 33.51 \\cdot 3.1416 \\approx \\boxed{105.19}A\u224833.51\u22c5\u03c0\u224833.51\u22c53.1416\u2248105.19\u200b<\/p>\n\n\n\n

Wait \u2014 we have an inconsistency. Let’s recheck:
Our earlier simplification was incorrect<\/strong> after taking the square root of that ratio.<\/p>\n\n\n\n

Let\u2019s back up and numerically evaluate the correct integral<\/strong>:<\/p>\n\n\n\n

\n\n\n\n

Correct Final Setup (for numerical integration):<\/h3>\n\n\n\n

A=2\u03c0\u222b0216\u22128y\u22c51+1616\u22128y\u2009dy=2\u03c0\u222b0216\u22128y+16\u2009dy=2\u03c0\u222b0232\u22128y\u2009dyA = 2\\pi \\int_0^2 \\sqrt{16 – 8y} \\cdot \\sqrt{1 + \\frac{16}{16 – 8y}} \\, dy = 2\\pi \\int_0^2 \\sqrt{16 – 8y + 16} \\, dy = 2\\pi \\int_0^2 \\sqrt{32 – 8y} \\, dyA=2\u03c0\u222b02\u200b16\u22128y\u200b\u22c51+16\u22128y16\u200b\u200bdy=2\u03c0\u222b02\u200b16\u22128y+16\u200bdy=2\u03c0\u222b02\u200b32\u22128y\u200bdy<\/p>\n\n\n\n

Let\u2019s integrate:\u222b0232\u22128y\u2009dy=\u221218\u22c5\u222b3216u\u2009du=\u221218\u22c5[23u3\/2]3216=\u221218\u22c523(163\/2\u2212323\/2)=\u221218\u22c523(64\u2212181.02)=\u221218\u22c523(\u2212117.02)=18\u22c523\u22c5117.02=234.0424=9.75Now:\\[A=2\u03c0\u22c59.75\u224861.27\\int_0^2 \\sqrt{32 – 8y} \\, dy = \\frac{-1}{8} \\cdot \\int_{32}^{16} \\sqrt{u} \\, du = \\frac{-1}{8} \\cdot \\left[ \\frac{2}{3} u^{3\/2} \\right]_{32}^{16} = \\frac{-1}{8} \\cdot \\frac{2}{3} (16^{3\/2} – 32^{3\/2}) = \\frac{-1}{8} \\cdot \\frac{2}{3} (64 – 181.02) = \\frac{-1}{8} \\cdot \\frac{2}{3} (-117.02) = \\frac{1}{8} \\cdot \\frac{2}{3} \\cdot 117.02 = \\frac{234.04}{24} = 9.75 Now: \\[ A = 2\\pi \\cdot 9.75 \\approx 61.27\u222b02\u200b32\u22128y\u200bdy=8\u22121\u200b\u22c5\u222b3216\u200bu\u200bdu=8\u22121\u200b\u22c5[32\u200bu3\/2]3216\u200b=8\u22121\u200b\u22c532\u200b(163\/2\u2212323\/2)=8\u22121\u200b\u22c532\u200b(64\u2212181.02)=8\u22121\u200b\u22c532\u200b(\u2212117.02)=81\u200b\u22c532\u200b\u22c5117.02=24234.04\u200b=9.75Now:\\[A=2\u03c0\u22c59.75\u224861.27<\/p>\n\n\n\n

\n\n\n\n

\u2705 Final Answer: 61.27<\/strong><\/h3>\n\n\n\n
\n\n\n\n

\ud83d\udcd8 Textbook-style Explanation:<\/h3>\n\n\n\n

To find the surface area generated by rotating a curve around the y-axis, we use the formulaA=2\u03c0\u222babx1+(dxdy)2\u2009dyA = 2\\pi \\int_{a}^{b} x \\sqrt{1 + \\left(\\frac{dx}{dy}\\right)^2} \\, dyA=2\u03c0\u222bab\u200bx1+(dydx\u200b)2\u200bdy<\/p>\n\n\n\n

For the curve x2=16\u22128yx^2 = 16 – 8yx2=16\u22128y, we solve for xxx and get x=16\u22128yx = \\sqrt{16 – 8y}x=16\u22128y\u200b.
Differentiating with respect to yyy, we find dxdy=\u2212416\u22128y\\frac{dx}{dy} = \\frac{-4}{\\sqrt{16 – 8y}}dydx\u200b=16\u22128y\u200b\u22124\u200b.
This leads to the surface area integral:A=2\u03c0\u222b0216\u22128y\u22c51+1616\u22128y\u2009dy=2\u03c0\u222b0232\u22128y\u2009dyA = 2\\pi \\int_0^2 \\sqrt{16 – 8y} \\cdot \\sqrt{1 + \\frac{16}{16 – 8y}} \\, dy = 2\\pi \\int_0^2 \\sqrt{32 – 8y} \\, dyA=2\u03c0\u222b02\u200b16\u22128y\u200b\u22c51+16\u22128y16\u200b\u200bdy=2\u03c0\u222b02\u200b32\u22128y\u200bdy<\/p>\n\n\n\n

With a substitution and evaluation, we find the surface area is approximately61.27 square units\\boxed{61.27 \\text{ square units}}61.27 square units\u200b<\/p>\n\n\n\n

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

Find the surface area generated by rotating the first quadrant portion of the curve x\\textsuperscript{2}=16-8y about the y-axis. ? 64.25 ? 66.38 ? 61.27 ? 58.41 The Correct Answer and Explanation is: To find the surface area generated by rotating the first-quadrant portion of the curvex2=16\u22128yx^2 = 16 – 8yx2=16\u22128y about the y-axis, we use […]<\/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-233487","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233487","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=233487"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233487\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233487"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233487"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233487"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}