{"id":717,"date":"2025-10-09T15:34:18","date_gmt":"2025-10-09T15:34:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=717"},"modified":"2025-12-12T15:01:35","modified_gmt":"2025-12-12T15:01:35","slug":"alternative-coordinate-systems-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/alternative-coordinate-systems-apply-it\/","title":{"raw":"Triple Integrals in Alternative Coordinate Systems: Apply It","rendered":"Triple Integrals in Alternative Coordinate Systems: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Evaluate a triple integral by changing to cylindrical coordinates<\/li>\r\n \t<li>Evaluate a triple integral by changing to spherical coordinates<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p id=\"fs-id1167793285292\">Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over [latex]500[\/latex] balloons participating each year.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1403\" align=\"aligncenter\" width=\"900\"]<img class=\"size-full wp-image-1403\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174048\/5-5-12.jpeg\" alt=\"A picture of many hot air balloons.\" width=\"900\" height=\"600\" \/> Figure 8.\u00a0Balloons lift off at the 2001 Albuquerque International Balloon Fiesta. (credit: David Herrera, Flickr)[\/caption]\r\n<p id=\"fs-id1167793510308\">As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however\u2014balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.<\/p>\r\n<p id=\"fs-id1167793510321\">In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius [latex]28[\/latex] feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is [latex]28[\/latex] feet and the radius of the small end of the frustum is [latex]6[\/latex] feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1404\" align=\"aligncenter\" width=\"801\"]<img class=\"size-full wp-image-1404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174119\/5-5-13.jpeg\" alt=\"This figure consists of two parts, a and b. Figure a shows a representation of a hot air balloon in xyz space as a half sphere on top of a frustrum of a cone. Figure b shows the dimensions, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, and the diameter of the circle at the top of the frustrum is 12 ft.\" width=\"801\" height=\"458\" \/> Figure 9.\u00a0(a) Use a half sphere to model the top part of the balloon and a frustum of a cone to model the bottom part of the balloon. (b) A cross section of the balloon showing its dimensions.[\/caption]\r\n<p id=\"fs-id1167794200502\">We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.<\/p>\r\n\r\n<ol id=\"fs-id1167794200510\" type=\"1\">\r\n \t<li>Find the volume of the balloon in two ways.\r\n<ol id=\"fs-id1167794200518\" type=\"a\">\r\n \t<li>Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)<\/li>\r\n \t<li>Verify the answer using the formulas for the volume of a sphere, [latex]V=\\frac43\\pi{r}^3[\/latex], and for the volume of a cone, [latex]V=\\frac13\\pi{r}^2h[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\nIn reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793937390\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<ol start=\"2\">\r\n \t<li>What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)<\/li>\r\n<\/ol>\r\nNow the pilot activates the burner for [latex]10[\/latex] seconds. This action affects the temperature in a [latex]12[\/latex]-foot-wide column [latex]20[\/latex] feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Calc_Figure_15_05_SP3\" class=\"os-figure\">\r\n<figure data-id=\"CNX_Calc_Figure_15_05_SP3\">[caption id=\"attachment_1405\" align=\"aligncenter\" width=\"385\"]<img class=\"size-full wp-image-1405\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174158\/5-5-14.jpeg\" alt=\"This figure shows the dimensions of the balloon and the hot air, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, the diameter of the circle at the top of the frustrum is 12 ft, and the inner column of hot air has height 20 ft and diameter 12 ft.\" width=\"385\" height=\"385\" \/> Figure 10.\u00a0<span class=\"os-caption\">Activating the burner heats the air in a [latex]20[\/latex]-foot-high, [latex]12[\/latex]-foot-wide column directly above the burner.<\/span>[\/caption]<\/figure>\r\n<\/div>\r\nAssume that after the pilot activates the burner for [latex]10[\/latex] seconds, the temperature of the air in the column described above\u00a0<em data-effect=\"italics\">increases<\/em>\u00a0according to the formula<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793376255\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{H(r,\\theta,z)=-2z-48.}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Then the temperature of the air in the column is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794067545\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{T_1(r,\\theta,z)=\\frac{z-r}{10}+210+(-2z-48)}[\/latex],<\/p>\r\n\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>while the temperature in the remainder of the balloon is still given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794292164\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<ol start=\"3\">\r\n \t<li>Find the average temperature of the air in the balloon after the pilot has activated the burner for [latex]10[\/latex] seconds.<\/li>\r\n<\/ol>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Evaluate a triple integral by changing to cylindrical coordinates<\/li>\n<li>Evaluate a triple integral by changing to spherical coordinates<\/li>\n<\/ul>\n<\/section>\n<p id=\"fs-id1167793285292\">Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over [latex]500[\/latex] balloons participating each year.<\/p>\n<figure id=\"attachment_1403\" aria-describedby=\"caption-attachment-1403\" style=\"width: 900px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1403\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174048\/5-5-12.jpeg\" alt=\"A picture of many hot air balloons.\" width=\"900\" height=\"600\" \/><figcaption id=\"caption-attachment-1403\" class=\"wp-caption-text\">Figure 8.\u00a0Balloons lift off at the 2001 Albuquerque International Balloon Fiesta. (credit: David Herrera, Flickr)<\/figcaption><\/figure>\n<p id=\"fs-id1167793510308\">As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however\u2014balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.<\/p>\n<p id=\"fs-id1167793510321\">In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius [latex]28[\/latex] feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is [latex]28[\/latex] feet and the radius of the small end of the frustum is [latex]6[\/latex] feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.<\/p>\n<figure id=\"attachment_1404\" aria-describedby=\"caption-attachment-1404\" style=\"width: 801px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174119\/5-5-13.jpeg\" alt=\"This figure consists of two parts, a and b. Figure a shows a representation of a hot air balloon in xyz space as a half sphere on top of a frustrum of a cone. Figure b shows the dimensions, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, and the diameter of the circle at the top of the frustrum is 12 ft.\" width=\"801\" height=\"458\" \/><figcaption id=\"caption-attachment-1404\" class=\"wp-caption-text\">Figure 9.\u00a0(a) Use a half sphere to model the top part of the balloon and a frustum of a cone to model the bottom part of the balloon. (b) A cross section of the balloon showing its dimensions.<\/figcaption><\/figure>\n<p id=\"fs-id1167794200502\">We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.<\/p>\n<ol id=\"fs-id1167794200510\" type=\"1\">\n<li>Find the volume of the balloon in two ways.\n<ol id=\"fs-id1167794200518\" type=\"a\">\n<li>Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)<\/li>\n<li>Verify the answer using the formulas for the volume of a sphere, [latex]V=\\frac43\\pi{r}^3[\/latex], and for the volume of a cone, [latex]V=\\frac13\\pi{r}^2h[\/latex].<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793937390\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex]<\/p>\n<\/div>\n<ol start=\"2\">\n<li>What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)<\/li>\n<\/ol>\n<p>Now the pilot activates the burner for [latex]10[\/latex] seconds. This action affects the temperature in a [latex]12[\/latex]-foot-wide column [latex]20[\/latex] feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Calc_Figure_15_05_SP3\" class=\"os-figure\">\n<figure data-id=\"CNX_Calc_Figure_15_05_SP3\">\n<figure id=\"attachment_1405\" aria-describedby=\"caption-attachment-1405\" style=\"width: 385px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1405\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174158\/5-5-14.jpeg\" alt=\"This figure shows the dimensions of the balloon and the hot air, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, the diameter of the circle at the top of the frustrum is 12 ft, and the inner column of hot air has height 20 ft and diameter 12 ft.\" width=\"385\" height=\"385\" \/><figcaption id=\"caption-attachment-1405\" class=\"wp-caption-text\">Figure 10.\u00a0<span class=\"os-caption\">Activating the burner heats the air in a [latex]20[\/latex]-foot-high, [latex]12[\/latex]-foot-wide column directly above the burner.<\/span><\/figcaption><\/figure>\n<\/figure>\n<\/div>\n<p>Assume that after the pilot activates the burner for [latex]10[\/latex] seconds, the temperature of the air in the column described above\u00a0<em data-effect=\"italics\">increases<\/em>\u00a0according to the formula<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793376255\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{H(r,\\theta,z)=-2z-48.}[\/latex]<\/p>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Then the temperature of the air in the column is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794067545\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{T_1(r,\\theta,z)=\\frac{z-r}{10}+210+(-2z-48)}[\/latex],<\/p>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>while the temperature in the remainder of the balloon is still given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794292164\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex].<\/p>\n<\/div>\n<ol start=\"3\">\n<li>Find the average temperature of the air in the balloon after the pilot has activated the burner for [latex]10[\/latex] seconds.<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":15,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-717","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/717","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/717\/revisions"}],"predecessor-version":[{"id":2781,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/717\/revisions\/2781"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/434"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/717\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=717"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=717"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=717"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=717"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}