Compute $int int(a^2-x^2) dx dy$ taken over half the circle $x^2+y^2=a^2$ in the positive quadrant.











up vote
1
down vote

favorite












Since it is given as positive quadrant, I took the limits as follows,
w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






integration







share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    Since it is given as positive quadrant, I took the limits as follows,
    w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
    I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






    integration







    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Since it is given as positive quadrant, I took the limits as follows,
      w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
      I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






      integration







      share|cite|improve this question















      Since it is given as positive quadrant, I took the limits as follows,
      w.r.t. x from 0 to a & w.r.t y from 0 to root of $$(a^2-x^2)$$.
      I proceeded with that & finally substituted $x=asin(theta)$. The answer which I got was $(3(a^4)π)/16$. But the actual answer is $(3a^4)/8$. I don't know where I am making a mistake.






      integration




      integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 14 hours ago









      David G. Stork

      8,95021232




      8,95021232










      asked 15 hours ago









      Renuka

      123




      123






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote













          The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






          share|cite|improve this answer






























            up vote
            0
            down vote













            $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






            share|cite|improve this answer























              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',
              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%2f3004727%2fcompute-int-inta2-x2-dx-dy-taken-over-half-the-circle-x2y2-a2-in%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








              up vote
              1
              down vote













              The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






              share|cite|improve this answer



























                up vote
                1
                down vote













                The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.






                  share|cite|improve this answer














                  The question refers to a half circle and you are integrating over a quarter of a circle. Perhaps the intended region is ${(x,y):x geq 0, x^{2}+y^{2} leq 1$ in which case $y$ will vary from $-sqrt {a^{2}-x^{2}}$ to $sqrt {a^{2}-x^{2}}$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 14 hours ago

























                  answered 15 hours ago









                  Kavi Rama Murthy

                  40k31750




                  40k31750






















                      up vote
                      0
                      down vote













                      $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






                      share|cite|improve this answer



























                        up vote
                        0
                        down vote













                        $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$






                          share|cite|improve this answer














                          $$intlimits_{x=0}^a intlimits_{y=0}^sqrt{a^2-x^2} (a^2 -x^2) dy dx = frac{3 pi a^4}{16}$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 14 hours ago

























                          answered 14 hours ago









                          David G. Stork

                          8,95021232




                          8,95021232






























                               

                              draft saved


                              draft discarded



















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004727%2fcompute-int-inta2-x2-dx-dy-taken-over-half-the-circle-x2y2-a2-in%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

                              'app-layout' is not a known element: how to share Component with different Modules

                              Image
                              0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
                              Read more

                              android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
                              Read more

                              WPF add header to Image with URL pettitions [duplicate]

                              Image
                              up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
                              Read more