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- Thread starter Arun
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- Thread starter
- #1

- Feb 13, 2012

- 1,704

Wellcome on MHB Arun!...I want the answer for this and how is it solved.

double integral(x^{2}+y^{2}dxdy) over the region in pos quadrant for which x+y<=1.

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

Kind regards

$\chi$ $\sigma$

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- Jan 26, 2012

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Hi Arun,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.

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

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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.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

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- #6

- Jan 26, 2012

- 4,055

We use Latex on MHB and have a subforum 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.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.

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

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

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this is the imageWe use Latex on MHB and have a subforum 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:

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

- Feb 13, 2012

- 1,704

The simmetry 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

$\chi$ $\sigma$

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- #9

- Feb 21, 2013

- 739

Hello Arun,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.

I get it also to \(\displaystyle \frac{5}{12}\) 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 \(\displaystyle 1-x\) insted of \(\displaystyle 1-y\)

Regards,

\(\displaystyle |\pi\rangle\)

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