Solving an integral in polar coordinates











up vote
0
down vote

favorite
1












I am given the potential:



$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$



Where a is a constant.



I have to compute:



$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$



Changing to polar coordinates:



$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$



What I have done is:



$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$



Where



$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$



Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.



EDIT



Using polar coordinates here is not the best method.



Take squares and try:



$$x = u + v$$



$$y = u - v$$










share|cite|improve this question




















  • 1




    Is the integral meant to be indefinite?
    – J.G.
    Nov 22 at 11:55












  • @J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
    – JD_PM
    Nov 22 at 12:04










  • See math.stackexchange.com/questions/1096793/…
    – JD_PM
    Nov 23 at 15:34















up vote
0
down vote

favorite
1












I am given the potential:



$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$



Where a is a constant.



I have to compute:



$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$



Changing to polar coordinates:



$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$



What I have done is:



$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$



Where



$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$



Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.



EDIT



Using polar coordinates here is not the best method.



Take squares and try:



$$x = u + v$$



$$y = u - v$$










share|cite|improve this question




















  • 1




    Is the integral meant to be indefinite?
    – J.G.
    Nov 22 at 11:55












  • @J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
    – JD_PM
    Nov 22 at 12:04










  • See math.stackexchange.com/questions/1096793/…
    – JD_PM
    Nov 23 at 15:34













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I am given the potential:



$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$



Where a is a constant.



I have to compute:



$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$



Changing to polar coordinates:



$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$



What I have done is:



$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$



Where



$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$



Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.



EDIT



Using polar coordinates here is not the best method.



Take squares and try:



$$x = u + v$$



$$y = u - v$$










share|cite|improve this question















I am given the potential:



$$phi (x, y) = frac{k}{2}(x^2+y^2) + axy$$



Where a is a constant.



I have to compute:



$$int_{-infty}^infty int_{-infty}^infty mathrm e^{-betaphi}dxdy$$



Changing to polar coordinates:



$$int_{0}^{2pi}int_{0}^{b} e^{-betaphi}rdrdtheta$$



What I have done is:



$$iint e^{-betaphi}rdrdtheta = iint e^{-frac{beta Kr^2}{2}} e^{-beta a r^2 costheta sintheta}rdrdtheta = iint e^{-beta f(r, theta)}rdrdtheta$$



Where



$$f(r, theta) = frac{Kr^2}{2} [(1+frac{cos theta sin theta}{K})^2 -(frac{cos theta sin theta}{K})^2]$$



Now I guess I have to do a substitution, defining a new variable. But I do not get the right one, as I do not get a simpler form.



EDIT



Using polar coordinates here is not the best method.



Take squares and try:



$$x = u + v$$



$$y = u - v$$







calculus integration polar-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 15:33

























asked Nov 22 at 11:36









JD_PM

10410




10410








  • 1




    Is the integral meant to be indefinite?
    – J.G.
    Nov 22 at 11:55












  • @J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
    – JD_PM
    Nov 22 at 12:04










  • See math.stackexchange.com/questions/1096793/…
    – JD_PM
    Nov 23 at 15:34














  • 1




    Is the integral meant to be indefinite?
    – J.G.
    Nov 22 at 11:55












  • @J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
    – JD_PM
    Nov 22 at 12:04










  • See math.stackexchange.com/questions/1096793/…
    – JD_PM
    Nov 23 at 15:34








1




1




Is the integral meant to be indefinite?
– J.G.
Nov 22 at 11:55






Is the integral meant to be indefinite?
– J.G.
Nov 22 at 11:55














@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 at 12:04




@J.G. I'm not provided with the extremes but I guessed them. Note that $b$ is the distance with respect to a reference point.
– JD_PM
Nov 22 at 12:04












See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 at 15:34




See math.stackexchange.com/questions/1096793/…
– JD_PM
Nov 23 at 15:34















active

oldest

votes











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%2f3009030%2fsolving-an-integral-in-polar-coordinates%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3009030%2fsolving-an-integral-in-polar-coordinates%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

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei