Fdtxjr

Where does this common spurious transmission come from? Is there a quality difference?

Calculator final project in Python

Proper way to express "He disappeared them"

RigExpert AA-35 - Interpreting The Information

Running a General Election and the European Elections together

A Man With a Stainless Steel Endoskeleton (like The Terminator) Fighting Cloaked Aliens Only He Can See

WOW air has ceased operation, can I get my tickets refunded?

Why does standard notation not preserve intervals (visually)

Can you be charged for obstruction for refusing to answer questions?

How to prove a simple equation?

How is this set of matrices closed under multiplication?

Why did CATV standarize in 75 ohms and everyone else in 50?

How did people program for Consoles with multiple CPUs?

Reference request: Grassmannian and Plucker coordinates in type B, C, D

Method for adding error messages to a dictionary given a key

What flight has the highest ratio of timezone difference to flight time?

Is there always a complete, orthogonal set of unitary matrices?

Bartok - Syncopation (1): Meaning of notes in between Grand Staff

Why do remote US companies require working in the US?

Is wanting to ask what to write an indication that you need to change your story?

A small doubt about the dominated convergence theorem

Prepend last line of stdin to entire stdin

What did we know about the Kessel run before the prequels?

The exact meaning of 'Mom made me a sandwich'

Using spherical coordinates for triple integral

1

$begingroup$

$ int_0^3 int_0^sqrt9-x^2 int_0^sqrt9-x^2-y^2 fracsqrtx^2+y^2+z^21+x^2+y^2+z^2 dz dy dx$

Using spherical co-ord's this becomes :

$ int_0^2pi int_0^pi int_0^3 fracr^3sin(theta)1+r^2 dr dtheta dphi $

Is this correct? If it is how do I carry on from here?

integration multivariable-calculus spherical-coordinates

share|cite|improve this question

edited Mar 19 at 13:02

José Carlos Santos

171k23132240

asked Mar 19 at 11:49

Elliot SilverElliot Silver

406

$endgroup$

add a comment | 

1

$begingroup$

$ int_0^3 int_0^sqrt9-x^2 int_0^sqrt9-x^2-y^2 fracsqrtx^2+y^2+z^21+x^2+y^2+z^2 dz dy dx$

Using spherical co-ord's this becomes :

$ int_0^2pi int_0^pi int_0^3 fracr^3sin(theta)1+r^2 dr dtheta dphi $

Is this correct? If it is how do I carry on from here?

integration multivariable-calculus spherical-coordinates

share|cite|improve this question

edited Mar 19 at 13:02

José Carlos Santos

171k23132240

asked Mar 19 at 11:49

Elliot SilverElliot Silver

406

$endgroup$

add a comment | 

$begingroup$

$ int_0^3 int_0^sqrt9-x^2 int_0^sqrt9-x^2-y^2 fracsqrtx^2+y^2+z^21+x^2+y^2+z^2 dz dy dx$

Using spherical co-ord's this becomes :

$ int_0^2pi int_0^pi int_0^3 fracr^3sin(theta)1+r^2 dr dtheta dphi $

Is this correct? If it is how do I carry on from here?

integration multivariable-calculus spherical-coordinates

share|cite|improve this question

edited Mar 19 at 13:02

José Carlos Santos

171k23132240

asked Mar 19 at 11:49

Elliot SilverElliot Silver

406

$endgroup$

share|cite|improve this question

edited Mar 19 at 13:02

José Carlos Santos

171k23132240

asked Mar 19 at 11:49

Elliot SilverElliot Silver

406

share|cite|improve this question

edited Mar 19 at 13:02

José Carlos Santos

171k23132240

asked Mar 19 at 11:49

Elliot SilverElliot Silver

406

edited Mar 19 at 13:02

José Carlos Santos

171k23132240

edited Mar 19 at 13:02

José Carlos Santos

171k23132240

asked Mar 19 at 11:49

Elliot SilverElliot Silver

406

asked Mar 19 at 11:49

Elliot SilverElliot Silver

406

2 Answers
2

active

oldest

votes

1

$begingroup$

Your limits aren't quite correct. The spherical region you are integrating over is only in the first octant. Therefore $theta$ should range from 0 to $pi/2$ and $phi$ should range from $0$ to $pi/2$.

share|cite|improve this answer

answered Mar 19 at 12:59

SpencerSpencer

8,67012056

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, how can i show this using the given limits?
  $endgroup$
  – Elliot Silver
  Mar 19 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean how can you compute the integral now? Or do you mean how do you show that the limits on the angles are from $0$ to $pi/2$?
  $endgroup$
  – Spencer
  Mar 19 at 13:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've figured out how to get the limits now thank you, the integral i got in the end was pi/4( 9-ln(10)), could you tell me if this is correct?
  $endgroup$
  – Elliot Silver
  Mar 19 at 13:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That looks right to me. At this point you should probably accept the answer by José Carlos Santos.
  $endgroup$
  – Spencer
  Mar 19 at 13:17
  

  
  
  
  

add a comment | 

1

$begingroup$

It looks fine to me, except for the limits of integration. The next step is to write it as$$fracpi2left(int_0^fracpi2sin(theta),mathrm dthetaright)left(int_0^3fracr^31+r^2,mathrm drright).$$

share|cite|improve this answer

edited Mar 19 at 13:02

answered Mar 19 at 11:55

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just wondering if the limits for the spherical were correct?
  $endgroup$
  – Elliot Silver
  Mar 19 at 12:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No. I did not notice that you are working on the first octant only. Therefore, both $phi$ and $theta$ shoulg go only from $0$ to $fracpi2$. I shall edit my answer.
  $endgroup$
  – José Carlos Santos
  Mar 19 at 13:01
  

  
  
  
  

add a comment | 

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

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3153966%2fusing-spherical-coordinates-for-triple-integral%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Your limits aren't quite correct. The spherical region you are integrating over is only in the first octant. Therefore $theta$ should range from 0 to $pi/2$ and $phi$ should range from $0$ to $pi/2$.

share|cite|improve this answer

answered Mar 19 at 12:59

SpencerSpencer

8,67012056

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, how can i show this using the given limits?
  $endgroup$
  – Elliot Silver
  Mar 19 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean how can you compute the integral now? Or do you mean how do you show that the limits on the angles are from $0$ to $pi/2$?
  $endgroup$
  – Spencer
  Mar 19 at 13:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've figured out how to get the limits now thank you, the integral i got in the end was pi/4( 9-ln(10)), could you tell me if this is correct?
  $endgroup$
  – Elliot Silver
  Mar 19 at 13:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That looks right to me. At this point you should probably accept the answer by José Carlos Santos.
  $endgroup$
  – Spencer
  Mar 19 at 13:17
  

  
  
  
  

add a comment | 

1

$begingroup$

Your limits aren't quite correct. The spherical region you are integrating over is only in the first octant. Therefore $theta$ should range from 0 to $pi/2$ and $phi$ should range from $0$ to $pi/2$.

share|cite|improve this answer

answered Mar 19 at 12:59

SpencerSpencer

8,67012056

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, how can i show this using the given limits?
  $endgroup$
  – Elliot Silver
  Mar 19 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean how can you compute the integral now? Or do you mean how do you show that the limits on the angles are from $0$ to $pi/2$?
  $endgroup$
  – Spencer
  Mar 19 at 13:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've figured out how to get the limits now thank you, the integral i got in the end was pi/4( 9-ln(10)), could you tell me if this is correct?
  $endgroup$
  – Elliot Silver
  Mar 19 at 13:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  That looks right to me. At this point you should probably accept the answer by José Carlos Santos.
  $endgroup$
  – Spencer
  Mar 19 at 13:17
  

  
  
  
  

add a comment | 

$begingroup$

Your limits aren't quite correct. The spherical region you are integrating over is only in the first octant. Therefore $theta$ should range from 0 to $pi/2$ and $phi$ should range from $0$ to $pi/2$.

share|cite|improve this answer

answered Mar 19 at 12:59

SpencerSpencer

8,67012056

$endgroup$

share|cite|improve this answer

answered Mar 19 at 12:59

SpencerSpencer

8,67012056

answered Mar 19 at 12:59

SpencerSpencer

8,67012056

answered Mar 19 at 12:59

SpencerSpencer

8,67012056

1

$begingroup$

It looks fine to me, except for the limits of integration. The next step is to write it as$$fracpi2left(int_0^fracpi2sin(theta),mathrm dthetaright)left(int_0^3fracr^31+r^2,mathrm drright).$$

share|cite|improve this answer

edited Mar 19 at 13:02

answered Mar 19 at 11:55

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just wondering if the limits for the spherical were correct?
  $endgroup$
  – Elliot Silver
  Mar 19 at 12:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No. I did not notice that you are working on the first octant only. Therefore, both $phi$ and $theta$ shoulg go only from $0$ to $fracpi2$. I shall edit my answer.
  $endgroup$
  – José Carlos Santos
  Mar 19 at 13:01
  

  
  
  
  

add a comment | 

1

$begingroup$

It looks fine to me, except for the limits of integration. The next step is to write it as$$fracpi2left(int_0^fracpi2sin(theta),mathrm dthetaright)left(int_0^3fracr^31+r^2,mathrm drright).$$

share|cite|improve this answer

edited Mar 19 at 13:02

answered Mar 19 at 11:55

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was just wondering if the limits for the spherical were correct?
  $endgroup$
  – Elliot Silver
  Mar 19 at 12:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No. I did not notice that you are working on the first octant only. Therefore, both $phi$ and $theta$ shoulg go only from $0$ to $fracpi2$. I shall edit my answer.
  $endgroup$
  – José Carlos Santos
  Mar 19 at 13:01
  

  
  
  
  

add a comment | 

$begingroup$

It looks fine to me, except for the limits of integration. The next step is to write it as$$fracpi2left(int_0^fracpi2sin(theta),mathrm dthetaright)left(int_0^3fracr^31+r^2,mathrm drright).$$

share|cite|improve this answer

edited Mar 19 at 13:02

answered Mar 19 at 11:55

José Carlos SantosJosé Carlos Santos

171k23132240

$endgroup$

share|cite|improve this answer

edited Mar 19 at 13:02

answered Mar 19 at 11:55

José Carlos SantosJosé Carlos Santos

171k23132240

answered Mar 19 at 11:55

José Carlos SantosJosé Carlos Santos

171k23132240

answered Mar 19 at 11:55

José Carlos SantosJosé Carlos Santos

171k23132240