{"id":20205,"date":"2024-04-15T05:49:47","date_gmt":"2024-04-15T05:49:47","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20205"},"modified":"2024-04-15T05:49:47","modified_gmt":"2024-04-15T05:49:47","slug":"the-area-enclosed-between-the-curves-y2-4x-and-x2-4y-is-a-frac163-b-8-c-frac323-d-16","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-area-enclosed-between-the-curves-y2-4x-and-x2-4y-is-a-frac163-b-8-c-frac323-d-16\/","title":{"rendered":"The area enclosed between the curves y2 = 4x and x2 = 4y is A. \\[\\frac{{16}}{3}\\] B. 8 C. \\[\\frac{{32}}{3}\\] D. 16"},"content":{"rendered":"

[amp_mcq option1=”\\[\\frac{{16}}{3}\\]” option2=”8″ option3=”\\[\\frac{{32}}{3}\\]” option4=”16″ correct=”option1″]<\/p>\n

The area enclosed between the curves $y^2 = 4x$ and $x^2 = 4y$ is $\\frac{32}{3}$ square units.<\/p>\n

To find the area, we can use the following steps:<\/p>\n

    \n
  1. Find the points of intersection of the two curves.<\/li>\n
  2. Set up a definite integral that represents the area enclosed by the curves.<\/li>\n
  3. Evaluate the definite integral.<\/li>\n<\/ol>\n

    The points of intersection of the two curves are $(0,0)$ and $(4,4)$.<\/p>\n

    The definite integral that represents the area enclosed by the curves is:<\/p>\n

    $$\\int_0^4 \\left( y^2 – x^2 \\right) \\, dx$$<\/p>\n

    Evaluating the definite integral, we get:<\/p>\n

    $$\\left[ \\frac{y^3}{3} – \\frac{x^3}{3} \\right]_0^4 = \\frac{64}{3} – \\frac{64}{3} = \\boxed{\\frac{32}{3}}$$<\/p>\n

    Here is a graph of the two curves:<\/p>\n

    [asy]
    \nunitsize(1 cm);<\/p>\n

    real ticklen=3;
    \nreal tickspace=1;
    \nreal axisarrowsize=0.14inch;
    \nreal tickdown=-0.5;
    \nreal tickdownlength=-0.12inch;
    \nreal wholetickdown=-0.24inch;
    \nreal wholetickdownlength=-0.2inch;
    \nreal tickdownlengthshort=-0.08inch;
    \nreal wholetickdownlengthshort=-0.12inch;
    \nreal t=0;
    \nreal d=1;
    \nreal xleft=-4;
    \nreal xright=4;
    \nreal ybottom=-4;
    \nreal ytop=4;<\/p>\n

    draw((xleft,0)–(xright,0));
    \ndraw((0,ybottom)–(0,ytop));<\/p>\n

    real i;<\/p>\n

    for(i=xleft+tickspace; i<xright; i+=tickspace) {
    \n draw((i,0.1)–(i,-0.1));
    \n}<\/p>\n

    for(i=ybottom+tickspace; i<ytop; i+=tickspace) {
    \n draw((0.1,i)–(-0.1,i));
    \n}<\/p>\n

    label(“$x$”,(xright+0.2,0),SE);
    \nlabel(“$y$”,(0,ytop+0.2),NW);<\/p>\n

    draw((0,-0.1)–(0,0.1));
    \ndraw((0.1,-0.1)–(0.1,0.1));<\/p>\n

    label(“$y^2=4x$”,(4,2),S);
    \nlabel(“$x^2=4y$”,(2,4),E);<\/p>\n

    draw(graph(y^2=4x,-4,4),red);
    \ndraw(graph(x^2=4y,-4,4),blue);
    \n[\/asy]<\/p>\n","protected":false},"excerpt":{"rendered":"

    [amp_mcq option1=”\\[\\frac{{16}}{3}\\]” option2=”8″ option3=”\\[\\frac{{32}}{3}\\]” option4=”16″ correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20205","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\nThe area enclosed between the curves y2 = 4x and x2 = 4y is A. \\[\\frac{{16}}{3}\\] B. 8 C. \\[\\frac{{32}}{3}\\] D. 16<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-area-enclosed-between-the-curves-y2-4x-and-x2-4y-is-a-frac163-b-8-c-frac323-d-16\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta 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