Solution to Double Integral Problem

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SUMMARY

The discussion centers on solving the double integral of the function \(x^2 + y^2\) over the region defined by \(x + y \leq 1\) in the positive quadrant. Participants confirm that the integral evaluates to \(\frac{5}{12}\), contrasting with a textbook answer of \(\frac{1}{6}\), suggesting a potential error in the text. The conversation emphasizes the importance of showing work for accurate troubleshooting and highlights the use of LaTeX for mathematical expressions.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with the concept of regions of integration
  • Basic knowledge of LaTeX for formatting mathematical expressions
  • Ability to perform algebraic manipulations and integration techniques
NEXT STEPS
  • Study the method of changing the order of integration in double integrals
  • Learn how to use LaTeX for mathematical documentation
  • Explore common pitfalls in evaluating double integrals
  • Review the properties of symmetric functions in integration
USEFUL FOR

Students and educators in mathematics, particularly those focusing on calculus and integral calculus, as well as anyone looking to improve their skills in presenting mathematical work using LaTeX.

TheArun
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I want the answer for this and how is it solved.
double integral(x2+y2 dxdy) over the region in pos quadrant for which x+y<=1.
 
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Re: iam new...i have a question...pls help.

Arun said:
I want the answer for this and how is it solved.
double integral(x2+y2 dxdy) over the region in pos quadrant for which x+y<=1.

Wellcome on MHB Arun!...

... did You make some attempt?...

Kind regards

$\chi$ $\sigma$
 
Re: iam new...i have a question...pls help.

ya...sure..tried many times...iam getting 2/3 as ans...but accd to text it is 1/6...dont know y iam wrong.
Btw thank for reply.
 
Re: iam new...i have a question...pls help.

Arun said:
ya...sure..tried many times...iam getting 2/3 as ans...but accd to text it is 1/6...dont know y iam wrong.
Btw thank for reply.

Hi Arun,

Welcome to MHB! :) Please use proper English here, meaning don't use lots of abbreviations like "ans" for "answer".

In order to help you we need to see what work you've done, so without seeing how you got $\frac{2}{3}$ we don't know where you went wrong.

Jameson
 
Re: iam new...i have a question...pls help.

Jameson said:
Hi Arun,

Welcome to MHB! :) Please use proper English here, meaning don't use lots of abbreviations like "ans" for "answer".

In order to help you we need to see what work you've done, so without seeing how you got $\frac{2}{3}$ we don't know where you went wrong.

Jameson

sorry fed up with all the subscripts and superscripts...is der some easier way...anyway i have done it in word.but it is showing invalid file.
 
Re: iam new...i have a question...pls help.

Arun said:
sorry fed up with all the subscripts and superscripts...is der some easier way...anyway i have done it in word.but it is showing invalid file.

We use Latex on MHB and have a http://www.mathhelpboards.com/f26/ that explains how to use it. Until you learn how to use Latex I suggest taking a screenshot of your work and uploading the image to TinyPic. Then you can post the picture here.

Here is an example of what $\LaTeX$ can do:

$$\int_0^{\infty}e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}$$
 
Re: iam new...i have a question...pls help.

Jameson said:
We use Latex on MHB and have a http://www.mathhelpboards.com/f26/ that explains how to use it. Until you learn how to use Latex I suggest taking a screenshot of your work and uploading the image to TinyPic. Then you can post the picture here.

Here is an example of what $\LaTeX$ can do:

$$\int_0^{\infty}e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}$$

this is the image
156966o.jpg
 
All right!... the region of integration is the 'colored area' of the figure...

http://www.123homepage.it/u/i69735807._szw380h285_.jpg.jfif

The symmetry respect to x and y is evident, so that we can choose one or the other order of integration and write...

$\displaystyle \int\int_{A} (x^{2} + y^{2})\ dy\ dx = \int_{0}^{1} dx\ \int_{0}^{1-x} (x^{2}+y^{2})\ dy = \int_{0}^{1} |x^{2}\ y + \frac{y^{3}}{3}|_{0}^{1-x}\ dx = \int_{0}^{1} (\frac{1}{3} - x + 2\ x^{2} - \frac{x^{3}}{3})\ dx$ (1)

Now are You able to proceed?...

Kind regards

$\chi$ $\sigma$
 
so it turns out to be 5/12 right...?
But answer in a text is shown to be 1/6...author's mistake is it?
And i would also like to know y u chose this order of integration is it easier this way...the logic?
A very big thanks.
 
  • #10
Arun said:
so it turns out to be 5/12 right...?
But answer in a text is shown to be 1/6...author's mistake is it?
And i would also like to know y u chose this order of integration is it easier this way...the logic?
A very big thanks.
Hello Arun,
I get it also to $$\frac{5}{12}$$ and about the order of integration both is same difficult/simple. You can try it out if you want.
edit: When you mean 'order of integration' I asume you mean why he did choose $$1-x$$ insted of $$1-y$$

Regards,
$$|\pi\rangle$$
 
Last edited:

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