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Integrate over area enclosed by curve

by sanitykey
Tags: curve, enclosed, integrate
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sanitykey
#1
Nov8-06, 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]
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OlderDan
#2
Nov8-06, 12:18 PM
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P: 3,033
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.
sanitykey
#3
Nov8-06, 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 :)

OlderDan
#4
Nov8-06, 07:45 PM
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P: 3,033
Integrate over area enclosed by curve

Quote Quote by sanitykey
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 :)
In any coordinates you use, dA multiplies the whole integrand. The r instead of r^2 was just from careless typing. Sorry. It should be r^2. I fixed it in my earlier post. Your latest expression looks good.
sanitykey
#5
Nov9-06, 08:31 AM
P: 93
Thanks again my main problem with these types of questions is visualising what it actually means.
sanitykey
#6
Nov9-06, 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]
OlderDan
#7
Nov9-06, 11:13 AM
Sci Advisor
HW Helper
P: 3,033
That looks good.


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