area of part of Archimedes's spiral

Multi tool use
Multi tool use












6















Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration







share|cite|improve this question


















  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












  • The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    – WarreG
    18 hours ago
















6















Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration







share|cite|improve this question


















  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












  • The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    – WarreG
    18 hours ago














6












6








6


1






Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration







share|cite|improve this question














Find the area of region inside the "first loop" of the Archimedes spiral (that is, the spiral for $0 le theta le 2pi$) and to the left of the $y$-axis.




The area the question wants is between $theta = pi/2$ and $theta = 3pi/2$ for the graph $r=theta$. Therefore, I computed the integral $int_{pi/2}^{3pi/2} theta ,dtheta = pi^2$. I even checked it with a graphing calculator to make sure that the integral was computed correctly. However, apparently, this is not a correct answer. Could someone help me see why?






calculus integration




calculus integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 18 hours ago









jjhhjjhh

2,09611121




2,09611121








  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












  • The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    – WarreG
    18 hours ago














  • 2




    Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
    – Shubham Johri
    18 hours ago








  • 2




    Isn't the area element $r dr dtheta$ in polar?
    – coffeemath
    18 hours ago












  • The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
    – WarreG
    18 hours ago








2




2




Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
– Shubham Johri
18 hours ago






Can you explain why you wrote the area as $inttheta dtheta$? Are you familiar with double integrals?
– Shubham Johri
18 hours ago






2




2




Isn't the area element $r dr dtheta$ in polar?
– coffeemath
18 hours ago






Isn't the area element $r dr dtheta$ in polar?
– coffeemath
18 hours ago














The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
– WarreG
18 hours ago




The integral is $int int r dtheta dr = frac{1}{2}int theta^2 dtheta$, you forgot to integrate $r$ first from zero to $theta$.
– WarreG
18 hours ago










2 Answers
2






active

oldest

votes


















7














Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



Graph






share|cite|improve this answer





























    5














    The integral for finding the area in polar coordinate is different from what you have.



    Please use the correct formula and you will get the correct answer.



    $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






    share|cite|improve this answer























    • An extra $pi$ there
      – Shubham Johri
      18 hours ago






    • 1




      Thanks, I fixed it, now it is correct.
      – Mohammad Riazi-Kermani
      18 hours ago











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068444%2farea-of-part-of-archimedess-spiral%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



    The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



    Graph






    share|cite|improve this answer


























      7














      Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



      The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



      Graph






      share|cite|improve this answer
























        7












        7








        7






        Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



        The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



        Graph






        share|cite|improve this answer












        Recall that the area element in polar coordinates is given by $r dr dtheta$. For any value of $theta,r$ ranges from $0totheta$. Further, $theta$ ranges from $pi/2to3pi/2$.



        The answer is $$int_{pi/2}^{3pi/2}int_0^theta r dr dtheta=frac{13pi^3}{24}$$



        Graph







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 18 hours ago









        Shubham JohriShubham Johri

        4,651717




        4,651717























            5














            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






            share|cite|improve this answer























            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago
















            5














            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






            share|cite|improve this answer























            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago














            5












            5








            5






            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $






            share|cite|improve this answer














            The integral for finding the area in polar coordinate is different from what you have.



            Please use the correct formula and you will get the correct answer.



            $$ A = (1/2) int r^2 dtheta $$ where in your case $ r=theta $







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 18 hours ago

























            answered 18 hours ago









            Mohammad Riazi-KermaniMohammad Riazi-Kermani

            41.4k42061




            41.4k42061












            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago


















            • An extra $pi$ there
              – Shubham Johri
              18 hours ago






            • 1




              Thanks, I fixed it, now it is correct.
              – Mohammad Riazi-Kermani
              18 hours ago
















            An extra $pi$ there
            – Shubham Johri
            18 hours ago




            An extra $pi$ there
            – Shubham Johri
            18 hours ago




            1




            1




            Thanks, I fixed it, now it is correct.
            – Mohammad Riazi-Kermani
            18 hours ago




            Thanks, I fixed it, now it is correct.
            – Mohammad Riazi-Kermani
            18 hours ago


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068444%2farea-of-part-of-archimedess-spiral%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            SW Tk31ReVW5mLAn,MfZ7h Hv9SgqNn,iw5QK6YlzeBK
            -
            gk6,i663RJ8562j1LslAKyCM,VodAWbVQ5oivTXZezmR56FbnP,uP8df4yyCfVjp

            Popular posts from this blog

            How to make a Squid Proxy server?

            Image
            1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari ...
            Read more

            Is this a new Fibonacci Identity?

            Image
            6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w...
            Read more

            19世紀

            Image
            この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料...
            Read more