Evaluate the integral by changing to cylindrical coordinates.<\/p>\n\n\n\n
Integral2-2 integralsquareroot 4 – y2-squareroot 4 – y2 integral2squareroot x2 + y2 xz dz dx dy<\/p>\n\n\n\n
integral3-3 integralsquareroot 9 – x20 integral9-x2-y20 squareroot x2 + y2 dz dy dx<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n We are given a triple integral, and the task is to evaluate it by converting to cylindrical coordinates<\/strong>. Let\u2019s first decode and understand the expression clearly.<\/p>\n\n\n\n The given expression seems to be:<\/p>\n\n\n\n $$ However, another version follows:<\/p>\n\n\n\n $$ We’ll evaluate the first integral<\/strong> by converting to cylindrical coordinates<\/strong>.<\/p>\n\n\n\n Cylindrical coordinates are:<\/p>\n\n\n\n In the integrand $xz$, we substitute:<\/p>\n\n\n\n $$ The limits:<\/p>\n\n\n\n Now rewrite the integral:<\/p>\n\n\n\n $$ Simplify the integrand:<\/p>\n\n\n\n $$ So the integral becomes:<\/p>\n\n\n\n $$ Innermost integral:<\/strong><\/p>\n\n\n\n $$ Middle integral:<\/strong><\/p>\n\n\n\n $$ Outer integral:<\/strong><\/p>\n\n\n\n $$ Since the outermost integral equals 0<\/strong>, the value of the entire triple integral is:<\/p>\n\n\n\n $$ This problem involves evaluating a triple integral using cylindrical coordinates. The original region of integration defines a circular disk<\/strong> of radius 2 in the xy-plane, and a height that depends on the distance from the origin, namely $z = 2\\sqrt{x^2 + y^2}$. Because the integrand contains $xz$, which becomes $r \\cos\\theta \\cdot z$ in cylindrical coordinates, we observe the presence of a trigonometric term, $\\cos\\theta$.<\/p>\n\n\n\n Cylindrical coordinates are useful in this case because the symmetry of the region (a disk) matches the natural structure of cylindrical systems. By expressing $x = r\\cos\\theta$, $y = r\\sin\\theta$, and $dx\\,dy = r\\,dr\\,d\\theta$, we transform the region into polar bounds, greatly simplifying the integration process.<\/p>\n\n\n\n After converting the triple integral, the integrand becomes $r^2 \\cos\\theta \\cdot z$, and the limits become $r \\in [0, 2]$, $\\theta \\in [0, 2\\pi]$, and $z \\in [0, 2r]$. Integrating with respect to $z$ gives $2r^2$. Continuing with respect to $r$, the integral becomes $\\frac{64}{5}$. However, when integrating the outermost function $\\cos\\theta$ from $0$ to $2\\pi$, we get zero because $\\cos\\theta$ is symmetric about the y-axis and cancels out over a full period.<\/p>\n\n\n\n This cancellation is key\u2014despite nonzero inner integrals, the final result is zero due to symmetry. Therefore, the value of the integral is:<\/p>\n\n\n\n $$ Evaluate the integral by changing to cylindrical coordinates. Integral2-2 integralsquareroot 4 – y2-squareroot 4 – y2 integral2squareroot x2 + y2 xz dz dx dy integral3-3 integralsquareroot 9 – x20 integral9-x2-y20 squareroot x2 + y2 dz dy dx The Correct Answer and Explanation is: We are given a triple integral, and the task is to evaluate […]<\/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-217095","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217095","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=217095"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217095\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217095"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217095"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217095"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nInterpreting the Integral:<\/h3>\n\n\n\n
\\int_{-2}^{2} \\int_{-\\sqrt{4 – y^2}}^{\\sqrt{4 – y^2}} \\int_{0}^{2\\sqrt{x^2 + y^2}} xz \\, dz \\, dx \\, dy
$$<\/p>\n\n\n\n
\\int_{-3}^{3} \\int_{-\\sqrt{9 – x^2}}^{\\sqrt{9 – x^2}} \\int_{0}^{\\sqrt{9 – x^2 – y^2}} \\sqrt{x^2 + y^2} \\, dz \\, dy \\, dx
$$<\/p>\n\n\n\n
\n\n\n\nStep 1: Change to Cylindrical Coordinates<\/h3>\n\n\n\n
\n
xz = (r \\cos\\theta)z
$$<\/p>\n\n\n\n
\n\n\n\nStep 2: Analyze the Region<\/h3>\n\n\n\n
\n
\u2192 Together define a disk of radius 2<\/strong> in the xy-plane.<\/li>\n\n\n\n
\u2192 This is $z \\in [0, 2r]$ in cylindrical coordinates.<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 3: Set up the Cylindrical Integral<\/h3>\n\n\n\n
\\int_{0}^{2\\pi} \\int_{0}^{2} \\int_{0}^{2r} (r \\cos\\theta)z \\cdot r \\, dz \\, dr \\, d\\theta
$$<\/p>\n\n\n\n
r^2 \\cos\\theta \\cdot z
$$<\/p>\n\n\n\n
\\int_{0}^{2\\pi} \\int_{0}^{2} \\int_{0}^{2r} r^2 \\cos\\theta \\cdot z \\, dz \\, dr \\, d\\theta
$$<\/p>\n\n\n\n
\n\n\n\nStep 4: Evaluate the Integral<\/h3>\n\n\n\n
\\int_{0}^{2r} z \\, dz = \\left[\\frac{1}{2}z^2\\right]_0^{2r} = \\frac{1}{2}(4r^2) = 2r^2
$$<\/p>\n\n\n\n
\\int_{0}^{2} r^2 \\cdot 2r^2 \\, dr = 2 \\int_{0}^{2} r^4 \\, dr = 2 \\cdot \\left[\\frac{r^5}{5}\\right]_0^2 = 2 \\cdot \\frac{32}{5} = \\frac{64}{5}
$$<\/p>\n\n\n\n
\\int_{0}^{2\\pi} \\cos\\theta \\, d\\theta = 0
$$<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\\boxed{0}
$$<\/p>\n\n\n\n
\n\n\n\n300-Word Explanation:<\/h3>\n\n\n\n
\\boxed{0}
$$<\/p>\n","protected":false},"excerpt":{"rendered":"