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Area of region

Arun

New member
May 10, 2013
5
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.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: iam new...i have a question...pls help.

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

$\chi$ $\sigma$
 

Arun

New member
May 10, 2013
5
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.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
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.
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
 

Arun

New member
May 10, 2013
5
Re: iam new...i have a question...pls help.

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.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Re: iam new...i have a question...pls help.

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 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:

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

Arun

New member
May 10, 2013
5
Re: iam new...i have a question...pls help.

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.

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

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

chisigma

Well-known member
Feb 13, 2012
1,704
All right!... the region of integration is the 'colored area' of the figure...



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

$\chi$ $\sigma$
 

Arun

New member
May 10, 2013
5
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.
 

Petrus

Well-known member
Feb 21, 2013
739
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 \(\displaystyle \frac{5}{12}\) 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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