It is currently Wed Dec 13, 2017 11:49 pm


All times are UTC [ DST ]




Post new topic Reply to topic  [ 7 posts ] 
Author Message
PostPosted: Thu Nov 16, 2017 1:46 pm 

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 7
Can you show me how to change the order of integration to the following integral (how to find the new limits of integration)?

$\int_{0}^{2}\int_{\sqrt{y}}^{1} (x^2+y^3x) dxdy$


Top
Offline Profile  
Reply with quote  

PostPosted: Thu Nov 16, 2017 2:28 pm 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 431
Location: Ioannina, Greece
The double integral is calculated over the closed region $D$ which can be represented as $$D=\big\{(x,y)\in\mathbb{R}\;|\; \sqrt{y}\leqslant {x}\leqslant 1, \; 0\leqslant {y}\leqslant 2\big\}\,.$$ Can you represent the same region $D$ in such way, such that the variable $x$ takes values from $0$ to $1$ and no from $\sqrt{y}$ to $1$?

(Try to draw $D$.)

_________________
Grigorios Kostakos


Top
Offline Profile  
Reply with quote  

PostPosted: Thu Nov 16, 2017 2:59 pm 

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 7
I tried drawing D but I got confused. I represent $x=\sqrt{y}$ as $y=x^2$ and plot the lines $y=2$, $x=1$. It seems to me that D is divided in two sections: $0\leq x\leq1$ , $0\leq y\leq x^2$ and $1\leq x\leq\sqrt{2}$ , $x^2\leq y\leq 2$ , because the curve intersects the vertical line. Is this possible?


Top
Offline Profile  
Reply with quote  

PostPosted: Thu Nov 16, 2017 3:59 pm 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 431
Location: Ioannina, Greece
andrew.tzeva wrote:
I tried drawing D but I got confused. I represent $x=\sqrt{y}$ as $y=x^2$ and plot the lines $y=2$, $x=1$. It seems to me that D is divided in two sections: $0\leq x\leq1$ , $0\leq y\leq x^2$ and $1\leq x\leq\sqrt{2}$ , $x^2\leq y\leq 2$ , because the curve intersects the vertical line. Is this possible?

Attachment:
region1.png



Andrew, I apologize. I wasn't careful and I consider $D$ to be the region $D_1$ in the scheme.
\[D=\big\{(x,y)\in\mathbb{R}\;|\; \sqrt{y}\leqslant {x}\leqslant 1, \; 0\leqslant {y}\leqslant 2\big\}\] is not a representation of $D_1\cup D_2$. $x$ must be $\leqslant 1$ and for $D_1\cup D_2$ this not the case.

So, must be a typo in the upper and lower limits of the integrals!

_________________
Grigorios Kostakos


Top
Offline Profile  
Reply with quote  

PostPosted: Thu Nov 16, 2017 4:23 pm 

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 7
Oh, I see...The example was taken from Marsden-Tromba's Vector Calculus. These books aren't the gospel truth after all!


Top
Offline Profile  
Reply with quote  

PostPosted: Thu Nov 16, 2017 4:49 pm 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 431
Location: Ioannina, Greece
andrew.tzeva wrote:
Oh, I see...The example was taken from Marsden-Tromba's Vector Calculus. These books aren't the gospel truth after all!
Marsden-Tromba's Vector Calculus is a good book (not "the gospel truth" for me) but even the masterpieces did not escape entirely from typos!

P.S. So, I suppose that your question about this double integral was answered (?)

_________________
Grigorios Kostakos


Top
Offline Profile  
Reply with quote  

PostPosted: Thu Nov 16, 2017 4:59 pm 

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 7
Yes, this typo confused me. Thank you again for everything you've done!


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 7 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net