Solve the line integral without Green's formula












1














I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.



$int_{L}{(2x-y)dx+(x-y)dy}$



$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $



The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.



I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.






integration multivariable-calculus line-integrals







share|cite|improve this question






















  • Tell your teacher that this is a sick way to define a $1$-chain.
    – Christian Blatter
    Nov 20 at 19:08
















1














I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.



$int_{L}{(2x-y)dx+(x-y)dy}$



$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $



The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.



I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.






integration multivariable-calculus line-integrals







share|cite|improve this question






















  • Tell your teacher that this is a sick way to define a $1$-chain.
    – Christian Blatter
    Nov 20 at 19:08














1












1








1







I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.



$int_{L}{(2x-y)dx+(x-y)dy}$



$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $



The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.



I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.






integration multivariable-calculus line-integrals







share|cite|improve this question













I'm starting on double and line integrals, and I'm stuck at this question.
It asks of me to calculate the following integral without using Green's theorem. Usually with Green's theorem I use polar coordinates easily, but I can't seem to make the same substitution here.



$int_{L}{(2x-y)dx+(x-y)dy}$



$L = {(x,y): x^{2}+y^{2}=2y, xgeq0}cup{(x,y):x^{2}+y^{2}=4, xleq0,ygeq0} $



The curve $L$ is oriented from the point $(0,0)$, that is counter-clockwise.



I am not sure which subsitution to use. I always used Greens theorem so far. How would I go about solving this without it? I thought of using some parameter $t$ but couldn't quite specify its range. Thank you.






integration multivariable-calculus line-integrals




integration multivariable-calculus line-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 17:32









math101

497




497












  • Tell your teacher that this is a sick way to define a $1$-chain.
    – Christian Blatter
    Nov 20 at 19:08


















  • Tell your teacher that this is a sick way to define a $1$-chain.
    – Christian Blatter
    Nov 20 at 19:08
















Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 at 19:08




Tell your teacher that this is a sick way to define a $1$-chain.
– Christian Blatter
Nov 20 at 19:08










1 Answer
1






active

oldest

votes


















1














You have two curves.



first part.



$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$



$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$



and the second part



$x = 2cos t\
y = 2sin t$



$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$



If you wanted to use greens.



Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.



Then



$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$






share|cite|improve this answer























  • Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
    – math101
    Nov 20 at 22:32










  • You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
    – Doug M
    Nov 20 at 22:42










  • So that's it! I see now , thanks!
    – math101
    Nov 20 at 23:13











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%2f3006630%2fsolve-the-line-integral-without-greens-formula%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














You have two curves.



first part.



$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$



$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$



and the second part



$x = 2cos t\
y = 2sin t$



$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$



If you wanted to use greens.



Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.



Then



$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$






share|cite|improve this answer























  • Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
    – math101
    Nov 20 at 22:32










  • You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
    – Doug M
    Nov 20 at 22:42










  • So that's it! I see now , thanks!
    – math101
    Nov 20 at 23:13
















1














You have two curves.



first part.



$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$



$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$



and the second part



$x = 2cos t\
y = 2sin t$



$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$



If you wanted to use greens.



Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.



Then



$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$






share|cite|improve this answer























  • Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
    – math101
    Nov 20 at 22:32










  • You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
    – Doug M
    Nov 20 at 22:42










  • So that's it! I see now , thanks!
    – math101
    Nov 20 at 23:13














1












1








1






You have two curves.



first part.



$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$



$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$



and the second part



$x = 2cos t\
y = 2sin t$



$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$



If you wanted to use greens.



Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.



Then



$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$






share|cite|improve this answer














You have two curves.



first part.



$x = 2sin tcos t = sin 2t\
y = 2sin^2 t = 1-cos 2t$



$int_limits{0}^{frac {pi}{2}} (2sin 2t + 1 - cos 2t )(2cos 2t)+(sin 2t +1-cos 2t)(2sin 2t) dt$



and the second part



$x = 2cos t\
y = 2sin t$



$int_limits{frac {pi}{2}}^{pi} (4cos t - 2sin t)(-2sin t)+(2cos t - sin t)(2cos t) dt$



If you wanted to use greens.



Then the you would add the straight line from $(-2,0)$ to $(0,0)$ to close the contour.



Then



$int_{C_1} P(x,y) dx + Q(x,y) dy + int_{C_2} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA\
int_{C_1} P(x,y) dx + Q(x,y) dy = iint frac {partial P}{partial y} - frac{partial Q}{partial x} dA- int_{C_2} P(x,y) dx + Q(x,y) dy$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 20 at 17:57

























answered Nov 20 at 17:51









Doug M

43.9k31854




43.9k31854












  • Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
    – math101
    Nov 20 at 22:32










  • You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
    – Doug M
    Nov 20 at 22:42










  • So that's it! I see now , thanks!
    – math101
    Nov 20 at 23:13


















  • Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
    – math101
    Nov 20 at 22:32










  • You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
    – Doug M
    Nov 20 at 22:42










  • So that's it! I see now , thanks!
    – math101
    Nov 20 at 23:13
















Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 at 22:32




Thanks, I understand everything except how you got the first parts parametric equations. Since the first part of the curve is the right side of a circle of radius 1 and y offset of 1, shouldn't y be $y=1+sin(t)$. And $x=cos(t)$.
– math101
Nov 20 at 22:32












You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 at 22:42




You could use that parameterization, it would be for a different range of t. That is fine. I said, $x = rcos t, y = rsin t$ plugging into $x^2 + y^2 = 2y$ gives $r = 2sin t$ And put that back in for $r$ in the equations of $x,y$
– Doug M
Nov 20 at 22:42












So that's it! I see now , thanks!
– math101
Nov 20 at 23:13




So that's it! I see now , thanks!
– math101
Nov 20 at 23:13


















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%2f3006630%2fsolve-the-line-integral-without-greens-formula%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

How to send String Array data to Server using php in android

Image
0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
Read more

Is anime1.com a legal site for watching anime?

Image
up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
Read more