Register to reply 
Integrate over area enclosed by curve 
Share this thread: 
#1
Nov806, 11:56 AM

P: 93

Hi, i've been asked to use polar coordinates and to integrate
[tex]\int\int_{A}^{} x^2 + y^2 + \sqrt{x^2 + y^2}dxdy[/tex] over the area A enclosed by the curve [tex]x^2 + y^2 = 4[/tex] I know [tex]x=rcos(\theta)[/tex] [tex]y=rsin(\theta)[/tex] [tex]dxdy=rdrd\theta[/tex] So i think the integral can be written as [tex]\int\int_{A}^{} r^3 \pm r^2 drd\theta[/tex] and i think the limits might be: [tex]\int_{\theta=0}^{2\pi}\int_{r=2}^{2} r^3 \pm r^2 drd\theta[/tex] But i think these are probably wrong and i'm not sure how to account for the negative areas and such because i can't use symmetry if it's a function like that can i? If i do that i think the area comes out as [tex]64\pi/3[/tex] 


#2
Nov806, 12:18 PM

Sci Advisor
HW Helper
P: 3,031

r is never negative in polar coordintes. The lower limit of the r integral is zero. The theta variable is what takes you all the way around the curve. When a square root is written as you have been given, the positve square root is implied. The integrand is just r^3 + r.



#3
Nov806, 12:33 PM

P: 93

Ok i think i understand would that mean it should be [tex]\int_{\theta=0}^{2\pi}\int_{r=0}^{2} r^3 + r^2 drd\theta[/tex]
Which i think comes out with an answer of [tex]40\pi/3[/tex] If the inetgrand is [tex]r^3 + r[/tex] does that mean the r from the [tex]rdrd\theta[/tex] is only multiplied to one of them? I wasn't sure when i was writing it and just guessed it'd be multiplied to both. Thanks for your reply btw :) 


#4
Nov806, 07:45 PM

Sci Advisor
HW Helper
P: 3,031

Integrate over area enclosed by curve



#5
Nov906, 08:31 AM

P: 93

Thanks again my main problem with these types of questions is visualising what it actually means.



#6
Nov906, 10:08 AM

P: 93

Didn't think i should post this in a different thread as it's sort of the same topic but could someone please check this for me:
Question: Integrate the integral below [tex]\int\sqrt{1+x^2}ds[/tex] along the curve [itex]y=\frac{x^2}{2}[/itex] between [tex]x=0[/tex] and [tex]x=1[/tex]. Remember that [itex]ds=\sqrt{dx^2 + dy^2}[/itex] What i did: [tex]\frac{dy}{dx}=x[/tex] => [tex]dy=xdx[/tex] => [tex]dy^2 = x^2 dx^2[/tex] [tex]\int\sqrt{1+x^2}\sqrt{dx^2 + dy^2}[/tex] => [tex]\int\sqrt{1+x^2}\sqrt{dx^2 + x^2 dx^2}[/tex] => [tex]\int\sqrt{1+x^2}\sqrt{dx^2 (1+x^2)}[/tex] => [tex]\int\sqrt{1+x^2}\sqrt{1+x^2}dx[/tex] => [tex]\int1+x^2dx[/tex] => [tex]\int_{0}^{1}1+x^2dx[/tex] => [tex][1+\frac{1^3}{3}] = \frac{4}{3}[/tex] 


#7
Nov906, 11:13 AM

Sci Advisor
HW Helper
P: 3,031

That looks good.



Register to reply 
Related Discussions  
Area enclosed by one petal of the rose...?  Calculus & Beyond Homework  1  
Area enclosed by two curves  Calculus  7  
Area enclosed by the curve  Calculus & Beyond Homework  2  
Area enclosed by sine and cosine.  Calculus & Beyond Homework  3  
Area enclosed by line and curve  integration  Calculus & Beyond Homework  2 