We are given a spherical tank of radius 6 meters, partially filled with fluid to a depth of half the radius<\/strong>. The density of the fluid<\/strong> varies with height hh (from the bottom of the tank) and is given by: \u03c1(h)=16\u2212h2(in kg\/m3)\\rho(h) = 16 – h^2 \\quad \\text{(in kg\/m}^3\\text{)}<\/p>\n\n\n\n We are asked to express the mass<\/strong> of the fluid using an integral of the form: \u222babf(h)\u2009dh\\int_a^b f(h) \\, dh<\/p>\n\n\n\n Since height is measured from the bottom of the tank<\/strong>, the fluid starts at h=0h = 0<\/strong>. The depth of the fluid is half the radius<\/strong>, so it fills up to: 62=3 meters from the bottom\\frac{6}{2} = 3 \\text{ meters from the bottom}<\/p>\n\n\n\n \u2705 Answer: 3<\/strong><\/p>\n\n\n\n To find total mass, we integrate density \u00d7 volume element<\/strong> over height hh.<\/p>\n\n\n\n At height hh, a horizontal slice of the fluid is a circular disk<\/strong>. The volume element is: dV=\u03c0r(h)2\u2009dhdV = \\pi r(h)^2 \\, dh<\/p>\n\n\n\n We need to find r(h)r(h), the radius of a horizontal slice at height hh, using geometry of the sphere.<\/p>\n\n\n\n The center of the sphere is at h=6h = 6. So the vertical distance from the center to height hh is \u2223h\u22126\u2223|h – 6|. By the Pythagorean Theorem: r(h)2=62\u2212(h\u22126)2=36\u2212(h\u22126)2r(h)^2 = 6^2 – (h – 6)^2 = 36 – (h – 6)^2<\/p>\n\n\n\n So, f(h)=\u03c1(h)\u22c5area of slice=(16\u2212h2)\u22c5\u03c0[36\u2212(h\u22126)2]f(h) = \\rho(h) \\cdot \\text{area of slice} = (16 – h^2) \\cdot \\pi [36 – (h – 6)^2]<\/p>\n\n\n\n \u2705 Answer:<\/strong> f(h)=(16\u2212h2)\u22c5\u03c0[36\u2212(h\u22126)2]f(h) = (16 – h^2) \\cdot \\pi [36 – (h – 6)^2]<\/p>\n\n\n\n Density is in kg\/m\u00b3, volume in m\u00b3, so mass is: kg\/m3\u00d7m3=kg\\text{kg\/m}^3 \\times \\text{m}^3 = \\text{kg}<\/p>\n\n\n\n \u2705 Answer: kilograms<\/strong><\/p>\n\n\n\n This problem involves determining the mass of a variable-density fluid inside a spherical tank using calculus. Since the tank is spherical with a radius of 6 meters and filled to half that radius (3 meters), we need to integrate over the vertical section of fluid from height 0 (bottom) to 3 meters (fluid surface). This sets the lower and upper limits of the integral: a=0a = 0 and b=3b = 3.<\/p>\n\n\n\n To compute the total mass, we use the principle: Mass=\u222bdensity\u00d7volume element\\text{Mass} = \\int \\text{density} \\times \\text{volume element}<\/p>\n\n\n\n At each height hh, the fluid forms a thin horizontal disk. Its volume is dV=\u03c0r(h)2\u2009dhdV = \\pi r(h)^2 \\, dh, where r(h)r(h) is the radius of the disk at height hh. Using geometry of a sphere centered at height 6 (since the radius is 6 m), the horizontal radius at height hh is: r(h)2=36\u2212(h\u22126)2r(h)^2 = 36 – (h – 6)^2<\/p>\n\n\n\n The density varies with height as \u03c1(h)=16\u2212h2\\rho(h) = 16 – h^2. Thus, the integrand f(h)f(h) becomes: f(h)=\u03c1(h)\u22c5\u03c0r(h)2=(16\u2212h2)\u22c5\u03c0[36\u2212(h\u22126)2]f(h) = \\rho(h) \\cdot \\pi r(h)^2 = (16 – h^2) \\cdot \\pi [36 – (h – 6)^2]<\/p>\n\n\n\n Multiplying density and cross-sectional area at each level gives mass of that slice. Integrating from 0 to 3 meters gives the total mass of fluid. Since density is in kg\/m\u00b3 and volume in m\u00b3, the resulting units are kilograms<\/strong>, which is the standard unit of mass in the SI system.<\/p>\n\n\n\n A particular tank is a sphere of radius 6 meters. Fluid to a depth of half the radius is in the tank. The density of the fluid is given by 8(h) = (16 \u2013 h2) kg\/m3 at a height h meters up from the bottom of the tank (so h = 0 is the bottom […]<\/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-218235","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218235","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=218235"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218235\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218235"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218235"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218235"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Lower limit of integration, aa:<\/h3>\n\n\n\n
\u2705 Answer: 0<\/strong><\/p>\n\n\n\n(b) Upper limit of integration, bb:<\/h3>\n\n\n\n
(c) Integrand, f(h)f(h):<\/h3>\n\n\n\n
(d) Units of mass:<\/h3>\n\n\n\n
\n\n\n\n<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-184.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"