{"id":4568,"date":"2013-06-29T06:47:54","date_gmt":"2013-06-29T06:47:54","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=4568"},"modified":"2026-03-25T11:40:19","modified_gmt":"2026-03-25T16:40:19","slug":"volume-and-surface-area-of-a-sphere","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/volume-and-surface-area-of-a-sphere\/","title":{"rendered":"Volume and Surface Area of a Sphere"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_Pa4NyWBQUyQ\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_Pa4NyWBQUyQ\" data-source-videoID=\"Pa4NyWBQUyQ\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Volume and Surface Area of a Sphere Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Volume and Surface Area of a Sphere\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_Pa4NyWBQUyQ:hover {cursor:pointer;} img#videoThumbnailImage_Pa4NyWBQUyQ {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1879-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_Pa4NyWBQUyQ\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_Pa4NyWBQUyQ\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Volume and Surface Area of a Sphere\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_Pa4NyWBQUyQ\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_Pa4NyWBQUyQ\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_Pa4NyWBQUyQ\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction p63_Function() {\n  var x = document.getElementById(\"p63\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"p63_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"p63\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Volume_of_a_Sphere\" class=\"smooth-scroll\">Volume of a Sphere<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Finding_the_Volume\" class=\"smooth-scroll\">Finding the Volume<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Finding_the_Surface_Area\" class=\"smooth-scroll\">Finding the Surface Area<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Volume_and_Surface_Area_of_a_Sphere_Practice_Questions\" class=\"smooth-scroll\">Volume and Surface Area of a Sphere Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label><a href=\"https:\/\/www.mometrix.com\/academy\/volume-of-a-sphere-calculator\/\" target=\"none\" style=\"margin: 0 auto;\"><span class=\"accordion_calculator_button\">Calculator<\/span><\/a><\/p>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Spheres are amazing. They are perfectly symmetrical. They\u2019re super strong because they have no weak points. They naturally occur in the form of tiny things like atoms, small things like water droplets and bubbles, and enormous things like our sun.<\/p>\n<p>And we can calculate the volume and the surface area of all of these objects by using relatively simple formulas. All we need to know is the <strong>radius<\/strong> of the sphere, which is the distance from the center of the sphere to any point on the surface of the sphere.<\/p>\n<h2><span id=\"Volume_of_a_Sphere\" class=\"m-toc-anchor\"><\/span>Volume of a Sphere<\/h2>\n<p>\nLet\u2019s tackle volume first. The <strong>volume<\/strong> of a sphere is the measure of how much space it takes up. We measure this in cubic units, such as cubic inches or cubic centimeters. We can picture these units as cubes that we could place inside the sphere to see how many we can fit. But since the sphere is curved, even millimeter-sized cubes wouldn\u2019t give an exact volume. It\u2019s like trying to make a sphere out of toy building blocks\u2014it never really looks like a real sphere no matter how small the blocks are. <\/p>\n<p>But fortunately, we have a formula to calculate the exact volume of a sphere:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/02\/sphere@72.webp\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" \/><\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(V=\\frac{4}{3}\\pi r^{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhere <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span> is the radius of the sphere. The formula isn\u2019t exactly new. The Greek philosopher Archimedes discovered it over 2,000 years ago. But it still works as well now as it did then. Probably even better since we have electronic calculators these days. <\/p>\n<h2><span id=\"Finding_the_Volume\" class=\"m-toc-anchor\"><\/span>Finding the Volume<\/h2>\n<p>\nSo let\u2019s find the volume of a common object\u2014a billiards ball used to play pool. <\/p>\n<p>An eight-ball is 5.7 centimeters in diameter. Remember that the radius of a circle is half the diameter, so all we have to do is divide 5.7 by 2 to find our radius, 2.85 centimeters. Now we can plug that into our formula: <\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(V=\\frac{4}{3}\\pi r^{3}\\)<br \/>\n\\(V=\\frac{4}{3}\\pi (2.85\\text{ cm})^{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSubstituting using parentheses is always a good idea to keep our order of operations clear.<\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(V=\\frac{4}{3}\\pi (23.149\\text { cm}^{3})\\)<br \/>\n\\(V=30.865\\pi \\text{ cm}^{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nLeaving our answer in terms of <span style=\"font-style:normal; font-size:90%\">\\(\\pi \\)<\/span> keeps our volume exact. When we multiply by <span style=\"font-style:normal; font-size:90%\">\\(\\pi \\)<\/span>, we\u2019re getting an approximate value, though a very, very close approximation: <\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(V\\approx 96.965\\text { cm}^{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our answer is a bit less than 100 cubic centimeters. Imagine trying to build an eight-ball-sized sphere out of 100 centimeter cubes.<\/p>\n<p>Not great! But it\u2019s 6 centimeters high and about the same volume. It seems we should be glad we don\u2019t have to build using small cubes to find the volume of spheres! <\/p>\n<h2><span id=\"Finding_the_Surface_Area\" class=\"m-toc-anchor\"><\/span>Finding the Surface Area<\/h2>\n<p>\nWe can also find the surface area of our eight ball. Unlike with prisms and solids, our sphere doesn\u2019t have any faces. Or maybe it has an infinite number of faces. Either way, we can&#8217;t find the surface area by adding up the face area. Instead, we find the area of its entire surface area all at once using the surface area formula for a sphere: <\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(SA=4\\pi r^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhere, again, <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span> is equal to our radius. Remember that the formula for the area of a two dimensional circle is <span style=\"font-style:normal; font-size:90%\">\\(A=\\pi r^{2}\\)<\/span>. The surface area of a sphere is simply 4 times that. We could try cutting out 4 billiard-ball-sized circles on centimeter graph paper and trying to paste them on a billiard ball, but we can probably imagine how well that would go! <\/p>\n<p>Let&#8217;s use our formula instead. The radius of our eight ball is 2.85 centimeters. So just like with volume, we simply plug that into our formula: <\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(SA=4\\pi r^{2}\\)<br \/>\n\\(SA=4\\pi (2.85 \\text{ cm})^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSubstituting using parentheses is always a good idea to keep our order of operations clear. <\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(SA=4\\pi (8.1225 \\text{ cm}^{2})\\)<br \/>\n\\(SA=32.49\\pi \\text{ cm}^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOnce again leaving our answer in terms of <span style=\"font-style:normal; font-size:90%\">\\(\\pi \\)<\/span> keeps our answer exact. Then we can find our very close approximation value by multiplying by <span style=\"font-style:normal; font-size:90%\">\\(\\pi \\)<\/span> on our calculator.<\/p>\n<div class=\"examplesentence\" style=\"font-size: 100%;\">\\(SA\\approx 102.07\\text{ cm}^{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur answer is a little bit more than 100 square centimeters. Remember that surface area is a measure of two-dimensional area so our units need to be square units just like the squares on our graph paper. <\/p>\n<p>So how can we use these formulas to find the volume and surface area of everyday objects? The key is finding the radius, but finding it directly is a little bit difficult because we would need to be able to measure from the center of the sphere. And the center of most spheres is difficult to locate and access! Fortunately, we can easily find the diameter of a spherical object like a ball, and then we can find the radius by calculating half the diameter. Then we can use our two formulas. And that&#8217;s all there is to it! <\/p>\n<p>I hope this video on finding the volume and surface area of spheres was helpful. Thanks for watching and happy studying!<\/p>\n<p>For more help, check out our <a class=\"ylist\" target=\"_blank\" rel=\"noopener noreferrer\" href=https:\/\/www.mometrix.com\/academy\/sphere-volume-calculator\/\">volume of a sphere calculator<\/a>!<\/p>\n<ul class=\"citelist\">\n<li><a href=\"https:\/\/www.mathopenref.com\/spherevolume.html\"target=\"_blank\">\u201cVolume of a Sphere &#8211; Math Open Reference.\u201d 2011. Mathopenref.com<\/a><\/li>\n<li><a href=\"https:\/\/www.mathsisfun.com\/geometry\/sphere.html\"target=\"_blank\">\u201cSphere.\u201d 2019. Mathsisfun.com<\/a><\/li>\n<li><a href=\"https:\/\/en.wikipedia.org\/wiki\/Billiard_ball\"target=\"_blank\">\u201cBilliard Ball.\u201d 2023. Wikipedia<\/a><\/li>\n<\/ul>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Volume_and_Surface_Area_of_a_Sphere_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Volume and Surface Area of a Sphere Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nFind the volume of a sphere that has a radius of 10 inches. Round your answer to the nearest hundredth.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(V\\approx 4{,}811.78\\text{ in}^3\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-2\">\\(V\\approx 4{,}188.79\\text{ in}^3\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(V\\approx 418.88\\text{ in}^3\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(V\\approx 41.87\\text{ in}^3\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the radius of the sphere, which is 10 inches. Substitute this value into the formula \\(V=\\frac{4}{3}r^3\\) to find the volume.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(r=10\\)<br \/>\n\\(V=\\frac{4}{3}\\pi r^3\\)<br \/>\n\\(V=(\\frac{4}{3})(\\pi)(10)^3\\)\n<\/p>\n<p>Next, simplify the expression 10<sup>3<\/sup>.<\/p>\n<p style=\"text-align: center;\">\\(V=(\\frac{4}{3})(\\pi)(1{,}000)\\)<\/p>\n<p>From here, simplify the equation by multiplying \\(\\frac{4}{3}\\) and 1,000.<\/p>\n<p style=\"text-align: center;\">\\(V=1333.\\overline{33}\\pi\\)<\/p>\n<p>Finally, multiply by \u03c0. Round your answer to the nearest hundredth place.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(V\\approx 4{,}188.7902\\)<br \/>\n\\(V\\approx 4{,}188.79\\text{ in}^3\\)\n<\/p>\n<p>The volume of the sphere is approximately \\(4{,}188.79\\) cubic inches.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nFind the surface area of a sphere that has a radius of six meters. Round your answer to the nearest hundredth. <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\(SA\\approx 4{,}523.89\\text{ m}^2\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(SA\\approx 150.72\\text{ m}^2\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(SA\\approx 126.61\\text{ m}^2\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\">\\(SA\\approx 452.39\\text{ m}^2\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the radius of the sphere, which is 6 meters. Substitute this value into the formula \\(SA=4\u03c0r^2\\) to find the surface area. <\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(r=6\\)<br \/>\n\\(SA=4\\pi r^2\\)<br \/>\n\\(SA=(4)(\\pi)(6)^2\\)\n<\/p>\n<p>Next, simplify the expression 6<sup>2<\/sup>.<\/p>\n<p style=\"text-align: center;\">\\(SA=(4)(\\pi)(36)\\)<\/p>\n<p>From here, simplify the equation by multiplying 4 and 36.<\/p>\n<p style=\"text-align: center;\">\\(SA=144\\pi\\)<\/p>\n<p>Finally, multiply by \\(\u03c0\\). Round your answer to the nearest hundredth place. <\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(SA\\approx 452.38934\\)<br \/>\n\\(SA\\approx 452.39\\text{ m}^2\\)<\/p>\n<p>The surface area of the sphere is approximately 452.39 square meters.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nFind the volume of a sphere that has a diameter of 4 feet. Round your answer to the nearest hundredth. <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(V\\approx 25.12\\text{ ft}^3\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-2\">\\(V\\approx 33.51\\text{ ft}^3\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(V\\approx 8.37\\text{ ft}^3\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(V\\approx 16.75\\text{ ft}^3\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the radius of the sphere. Since the radius of a sphere is half its diameter, divide 4 by 2 to get 2 feet. Substitute this value into the formula \\(V=\\frac{4}{3}\\pi r^3\\) to find the volume. <\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(r=\\frac{d}{2}=\\frac{4}{2}=2\\)<br \/>\n\\(V=\\frac{4}{3}\\pi r^3\\)<br \/>\n\\(V=(\\frac{4}{3})(\\pi)(2)^3\\)\n<\/p>\n<p>Next, simplify the expression 2<sup>3<\/sup>.<\/p>\n<p style=\"text-align: center;\">\\(V=\\frac{4}{3}(\\pi)(8)\\)<\/p>\n<p>From here, simplify the equation by multiplying \\(\\frac{4}{3}\\) and 8.<\/p>\n<p style=\"text-align: center;\">\\(V=10.\\overline{66}\\pi\\)<\/p>\n<p>Finally, multiply by \\(\\pi\\). Round your answer to the nearest hundredth place.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(V\\approx 33.51032\\)<br \/>\n\\(V\\approx 33.51\\text{ ft}^3\\)\n<\/p>\n<p>The volume of the sphere is approximately 33.51 cubic feet.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nJoel is watching the NBA All-Star Game and wants to know the volume of a basketball. He knows that the radius of an official NBA basketball is about 4.7 inches. Based on this information, find the basketball\u2019s volume. Round your answer to the nearest hundredth.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-4-1\">\\(V\\approx 434.89\\text{ in}^3\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\(V\\approx 29.45\\text{ in}^3\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\(V\\approx 77.87\\text{ in}^3\\)<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\(V\\approx 18.8\\text{ in}^3\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the radius of the sphere, which is 4.7 inches. Substitute this value into the formula \\(V=\\frac{4}{3}\u03c0r^3\\) to find the volume.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(r=4.7\\)<br \/>\n\\(V=\\frac{4}{3}\\pi r^3\\)<br \/>\n\\(V=(\\frac{4}{3})(\\pi)(4.7)^3\\)\n<\/p>\n<p>Next, simplify the expression \\(4.7^3\\).<\/p>\n<p style=\"text-align: center;\">\\(V=(\\frac{4}{3})(\\pi)(103.823)\\)<\/p>\n<p>From here, simplify the equation by multiplying \\(\\frac{4}{3}\\) and \\(103.823\\).<\/p>\n<p style=\"text-align: center;\">\\(V=138.430\\overline{66}\\pi\\)<\/p>\n<p>Finally, multiply by \\(\u03c0\\). Round your answer to the nearest hundredth place.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(V\\approx 434.89277\\)<br \/>\n\\(V\\approx 434.89\\text{ in}^3\\)\n<\/p>\n<p>The volume of the basketball is approximately 434.89 cubic inches.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nKate is wrapping a birthday present for her niece. She needs to figure out how much wrapping paper is needed to cover the surface area of a beach ball. The beach ball has a diameter of 14 inches. Based on this information, find the surface area of the beach ball. Round your answer to the nearest hundredth. <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">\\(SA\\approx 113.04\\text{ in}^2\\)<\/div><div class=\"PQ\"  id=\"PQ-5-2\">\\(SA\\approx 175.84\\text{ in}^2\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-3\">\\(SA\\approx 615.75\\text{ in}^2\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\(SA\\approx 1{,}077.02\\text{ in}^2\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the radius of the sphere. Since the radius of a sphere is half its diameter, divide 14 by 2 to get 7 inches. Substitute this value into the formula \\(SA=4\\pi r^2\\) to find the surface area. <\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(r=\\frac{d}{2}=\\frac{14}{2}=7\\)<br \/>\n\\(SA=4\\pi r^2\\)<br \/>\n\\(SA=(4)(\\pi)(7)^2\\)\n<\/p>\n<p>Next, simplify the expression 7<sup>2<\/sup>.<\/p>\n<p style=\"text-align: center;\">\\(SA=(4)(\\pi)(49)\\)<\/p>\n<p>From here, simplify the equation by multiplying 4 and 49.<\/p>\n<p style=\"text-align: center;\">\\(SA=196\\pi\\)<\/p>\n<p>Finally, multiply by \\(\\pi\\). Round your answer to the nearest hundredth place.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(SA\\approx 615.75216\\)<br \/>\n\\(SA\\approx 615.75\\text{ in}^2\\)<\/p>\n<p>The surface area of the sphere is approximately 615.75 square inches.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/geometry\/\">Return to Geometry Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Geometry Videos<\/p>\n","protected":false},"author":1,"featured_media":99724,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-4568","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-volume-and-surface-area","7":"page_type-video","8":"content_type-practice-questions","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4568","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=4568"}],"version-history":[{"count":7,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4568\/revisions"}],"predecessor-version":[{"id":261100,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/4568\/revisions\/261100"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/99724"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=4568"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}