{"id":222685,"date":"2025-05-31T15:27:48","date_gmt":"2025-05-31T15:27:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222685"},"modified":"2025-05-31T15:27:50","modified_gmt":"2025-05-31T15:27:50","slug":"compute-the-surface-integral-of-the-scalar-field-uxy-x-xy-y-over-the-surface-s-represented-by-a-square-with-x-a%cb%86%cb%86-01-and-y-a%cb%86%cb%86-01-located-on-the-plane-z-0","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/compute-the-surface-integral-of-the-scalar-field-uxy-x-xy-y-over-the-surface-s-represented-by-a-square-with-x-a%cb%86%cb%86-01-and-y-a%cb%86%cb%86-01-located-on-the-plane-z-0\/","title":{"rendered":"Compute the surface integral of the scalar field U(x,y) = (x-x’y-y) over the surface S represented by a square with x \u00e2\u02c6\u02c6 [0,1] and y \u00e2\u02c6\u02c6 [0,1] located on the plane z = 0."},"content":{"rendered":"\n

Compute the surface integral of the scalar field U(x,y) = (x-x’y-y) over the surface S represented by a square with x \u00e2\u02c6\u02c6 [0,1] and y \u00e2\u02c6\u02c6 [0,1] located on the plane z = 0.
Compute the surface integral of the scalar field U(x,y) = (x-x’y-y) over the surface S represented by a square with x \u00e2\u02c6\u02c6 [0,1] and y \u00e2\u02c6\u02c6 [0,1] located on the plane z = 0.<\/p>\n\n\n\n

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

We are given a scalar field: U(x,y)=x\u2212x\u2032y\u2212yU(x, y) = x – x’y – y<\/p>\n\n\n\n

and a surface SS defined as the unit square in the xyxy-plane, i.e., S:{(x,y,z)\u2223x\u2208[0,1],y\u2208[0,1],z=0}S: \\{(x, y, z) \\mid x \\in [0, 1], y \\in [0, 1], z = 0\\}<\/p>\n\n\n\n

The problem is to compute the surface integral of U(x,y)U(x, y) over this surface: \u222cSU(x,y)\u2009dS\\iint_S U(x, y)\\, dS<\/p>\n\n\n\n

Step 1: Interpret the scalar field<\/h3>\n\n\n\n

First, clarify the scalar field U(x,y)=x\u2212x\u2032y\u2212yU(x, y) = x – x’y – y. The variable x\u2032x’ is unclear; in mathematical notation, this often denotes a derivative or another variable. If this is a typo and x\u2032x’ actually means just xx, then: U(x,y)=x\u2212xy\u2212yU(x, y) = x – xy – y<\/p>\n\n\n\n

Assuming this interpretation, the function simplifies to: U(x,y)=x(1\u2212y)\u2212yU(x, y) = x(1 – y) – y<\/p>\n\n\n\n

Step 2: Set up the surface integral<\/h3>\n\n\n\n

Since the surface lies on the plane z=0z = 0, and dS=dx\u2009dydS = dx\\,dy, the surface integral becomes a double integral over the unit square: \u222cSU(x,y)\u2009dx\u2009dy=\u222b01\u222b01(x(1\u2212y)\u2212y)\u2009dx\u2009dy\\iint_S U(x, y)\\, dx\\,dy = \\int_0^1 \\int_0^1 \\left(x(1 – y) – y\\right)\\, dx\\,dy<\/p>\n\n\n\n

Step 3: Compute the integral<\/h3>\n\n\n\n

\u222b01\u222b01(x(1\u2212y)\u2212y)\u2009dx\u2009dy=\u222b01[\u222b01x(1\u2212y)\u2009dx\u2212\u222b01y\u2009dx]dy\\int_0^1 \\int_0^1 \\left(x(1 – y) – y\\right)\\, dx\\,dy = \\int_0^1 \\left[ \\int_0^1 x(1 – y) \\, dx – \\int_0^1 y \\, dx \\right] dy<\/p>\n\n\n\n

Evaluate the inner integrals:<\/p>\n\n\n\n

    \n
  • \u222b01x(1\u2212y)\u2009dx=(1\u2212y)\u222b01x\u2009dx=(1\u2212y)\u22c512=1\u2212y2\\int_0^1 x(1 – y)\\, dx = (1 – y) \\int_0^1 x\\, dx = (1 – y)\\cdot\\frac{1}{2} = \\frac{1 – y}{2}<\/li>\n\n\n\n
  • \u222b01y\u2009dx=y\u22c5(1\u22120)=y\\int_0^1 y\\, dx = y \\cdot (1 – 0) = y<\/li>\n<\/ul>\n\n\n\n

    Now evaluate the outer integral: \u222b01[1\u2212y2\u2212y]dy=\u222b01(1\u2212y\u22122y2)dy=\u222b01(1\u22123y2)dy=12\u222b01(1\u22123y)\u2009dy\\int_0^1 \\left[ \\frac{1 – y}{2} – y \\right] dy = \\int_0^1 \\left( \\frac{1 – y – 2y}{2} \\right) dy = \\int_0^1 \\left( \\frac{1 – 3y}{2} \\right) dy = \\frac{1}{2} \\int_0^1 (1 – 3y)\\, dy =12[y\u221232y2]01=12(1\u221232)=12\u22c5(\u221212)=\u221214= \\frac{1}{2} \\left[ y – \\frac{3}{2}y^2 \\right]_0^1 = \\frac{1}{2} \\left( 1 – \\frac{3}{2} \\right) = \\frac{1}{2} \\cdot \\left( -\\frac{1}{2} \\right) = -\\frac{1}{4}<\/p>\n\n\n\n

    \n\n\n\n

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

    \u221214\\boxed{-\\frac{1}{4}}<\/p>\n\n\n\n

    \n\n\n\n

    Explanation <\/h3>\n\n\n\n

    The given problem requires evaluating a surface integral of a scalar field<\/strong> over a flat region. Surface integrals of scalar fields represent the accumulation of a quantity (like temperature or density) over a surface area. In this case, the scalar field U(x,y)=x\u2212x\u2032y\u2212yU(x, y) = x – x’y – y is defined over the unit square in the xyxy-plane, where x,y\u2208[0,1]x, y \\in [0, 1].<\/p>\n\n\n\n

    A crucial step was interpreting the notation. The term x\u2032x’ is ambiguous\u2014often it denotes a derivative, but in this context, we assumed it’s a typographical error and interpreted it simply as xx. Thus, we rewrote the field as U(x,y)=x(1\u2212y)\u2212yU(x, y) = x(1 – y) – y, a more standard scalar expression.<\/p>\n\n\n\n

    Since the surface lies flat on the z=0z = 0 plane and spans the square [0,1]\u00d7[0,1][0,1] \\times [0,1], the surface integral reduces to a standard double integral<\/strong> over that square. That is, the surface element dSdS becomes dx\u2009dydx\\,dy, and we integrate the function directly over the 2D region.<\/p>\n\n\n\n

    The integral was computed step-by-step: first integrating with respect to xx, then with respect to yy. The result after simplification was: \u222cSU(x,y)\u2009dx\u2009dy=\u221214\\iint_S U(x, y)\\, dx\\,dy = -\\frac{1}{4}<\/p>\n\n\n\n

    This value represents the net accumulation<\/strong> of the scalar field over the unit square. The negative result indicates that, on average, the field has a net “negative contribution” over the square\u2014likely because the negative terms dominate over most of the domain.<\/p>\n\n\n\n

    This type of integral is fundamental in physics and engineering, where scalar fields describe quantities distributed over surfaces, such as pressure on a plate or temperature across a wall.<\/p>\n\n\n\n

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

    Compute the surface integral of the scalar field U(x,y) = (x-x’y-y) over the surface S represented by a square with x \u00e2\u02c6\u02c6 [0,1] and y \u00e2\u02c6\u02c6 [0,1] located on the plane z = 0.Compute the surface integral of the scalar field U(x,y) = (x-x’y-y) over the surface S represented by a square with x \u00e2\u02c6\u02c6 […]<\/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-222685","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222685","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=222685"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222685\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222685"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222685"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222685"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}