Area of triangle using double integrals












2












$begingroup$


I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
$$ int_0^t int_0^frac{t}{2} dudv$$
But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
What am I doing wrong here?
I need to calculate it this way, not with single integral, or geometrically.






integration multivariable-calculus area







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
    $$ int_0^t int_0^frac{t}{2} dudv$$
    But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
    What am I doing wrong here?
    I need to calculate it this way, not with single integral, or geometrically.






    integration multivariable-calculus area







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
      $$ int_0^t int_0^frac{t}{2} dudv$$
      But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
      What am I doing wrong here?
      I need to calculate it this way, not with single integral, or geometrically.






      integration multivariable-calculus area







      share|cite|improve this question











      $endgroup$




      I have one (rather simple) problem, but I'm stuck and can't figure out what I'm constantly doing wrong. I need to calculate area of triangle with points at $(0,0)$, $(t,0)$, $(t,frac{t}{2})$. In other words triangle under function $y=frac{x}{2}$, for $xin [0,t]$ I thought it is calculated with
      $$ int_0^t int_0^frac{t}{2} dudv$$
      But it turns out that this equals to $frac{t^2}{2}$, when obviously this area is $frac{ttimesfrac{t}{2}}{2} = frac{t^2}{4}$.
      What am I doing wrong here?
      I need to calculate it this way, not with single integral, or geometrically.






      integration multivariable-calculus area




      integration multivariable-calculus area






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 16:25









      José Carlos Santos

      153k22123225




      153k22123225










      asked Dec 4 '18 at 16:16









      Vasilije DavidovicVasilije Davidovic

      132




      132






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:32





















          0












          $begingroup$

          I think you made a mistake, the bounds on the inner integral should not be constant:
          $$
          int_0^t int_0^{v/2} dudv
          = int_0^t frac{v}{2} dv
          = left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your reply. That solved it
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:33











          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%2f3025762%2farea-of-triangle-using-double-integrals%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









          1












          $begingroup$

          The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:32


















          1












          $begingroup$

          The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:32
















          1












          1








          1





          $begingroup$

          The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$






          share|cite|improve this answer









          $endgroup$



          The integral that you actually computed corresponds to the area of a rectangle. You should actually compute$$int_0^tint_0^{frac x2},mathrm dy,mathrm dx.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 16:20









          José Carlos SantosJosé Carlos Santos

          153k22123225




          153k22123225












          • $begingroup$
            Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:32




















          • $begingroup$
            Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:32


















          $begingroup$
          Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
          $endgroup$
          – Vasilije Davidovic
          Dec 4 '18 at 16:32






          $begingroup$
          Silly me. Names for variables were different in my notes, so I got confused. Thank you very much for your reply.
          $endgroup$
          – Vasilije Davidovic
          Dec 4 '18 at 16:32













          0












          $begingroup$

          I think you made a mistake, the bounds on the inner integral should not be constant:
          $$
          int_0^t int_0^{v/2} dudv
          = int_0^t frac{v}{2} dv
          = left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your reply. That solved it
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:33
















          0












          $begingroup$

          I think you made a mistake, the bounds on the inner integral should not be constant:
          $$
          int_0^t int_0^{v/2} dudv
          = int_0^t frac{v}{2} dv
          = left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you very much for your reply. That solved it
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:33














          0












          0








          0





          $begingroup$

          I think you made a mistake, the bounds on the inner integral should not be constant:
          $$
          int_0^t int_0^{v/2} dudv
          = int_0^t frac{v}{2} dv
          = left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
          $$






          share|cite|improve this answer









          $endgroup$



          I think you made a mistake, the bounds on the inner integral should not be constant:
          $$
          int_0^t int_0^{v/2} dudv
          = int_0^t frac{v}{2} dv
          = left. frac{v^2}{4} right|_0^t = frac{t^2}{4}.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 16:20









          gt6989bgt6989b

          33.3k22452




          33.3k22452












          • $begingroup$
            Thank you very much for your reply. That solved it
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:33


















          • $begingroup$
            Thank you very much for your reply. That solved it
            $endgroup$
            – Vasilije Davidovic
            Dec 4 '18 at 16:33
















          $begingroup$
          Thank you very much for your reply. That solved it
          $endgroup$
          – Vasilije Davidovic
          Dec 4 '18 at 16:33




          $begingroup$
          Thank you very much for your reply. That solved it
          $endgroup$
          – Vasilije Davidovic
          Dec 4 '18 at 16:33


















          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%2f3025762%2farea-of-triangle-using-double-integrals%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







          -

          Popular posts from this blog

          To store a contact into the json file from server.js file using a class in NodeJS

          Image
          0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
          Read more

          Redirect URL with Chrome Remote Debugging Android Devices

          Image
          1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more