Integrate over area enclosed by curve

[sanitykey]

Hi, I've been asked to use polar coordinates and to integrate

\int\int_{A}^{} x^2 + y^2 + \sqrt{x^2 + y^2}dxdy

over the area A enclosed by the curve x^2 + y^2 = 4

I know

x=rcos(\theta)
y=rsin(\theta)
dxdy=rdrd\theta

So i think the integral can be written as

\int\int_{A}^{} r^3 \pm r^2 drd\theta

and i think the limits might be:

\int_{\theta=0}^{2\pi}\int_{r=-2}^{2} r^3 \pm r^2 drd\theta

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 64\pi/3

 

[OlderDan]

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]

Ok i think i understand would that mean it should be \int_{\theta=0}^{2\pi}\int_{r=0}^{2} r^3 + r^2 drd\theta


Which i think comes out with an answer of 40\pi/3

If the inetgrand is r^3 + r does that mean the r from the rdrd\theta 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]

Ok i think i understand would that mean it should be \int_{\theta=0}^{2\pi}\int_{r=0}^{2} r^3 + r^2 drd\theta


Which i think comes out with an answer of 40\pi/3

If the inetgrand is r^3 + r does that mean the r from the rdrd\theta 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]

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

 

[sanitykey]

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

\int\sqrt{1+x^2}ds

along the curve y=\frac{x^2}{2} between x=0 and x=1. Remember that ds=\sqrt{dx^2 + dy^2}

What i did:

\frac{dy}{dx}=x


=> dy=xdx


=> dy^2 = x^2 dx^2


\int\sqrt{1+x^2}\sqrt{dx^2 + dy^2}


=> \int\sqrt{1+x^2}\sqrt{dx^2 + x^2 dx^2}


=> \int\sqrt{1+x^2}\sqrt{dx^2 (1+x^2)}


=> \int\sqrt{1+x^2}\sqrt{1+x^2}dx


=> \int1+x^2dx


=> \int_{0}^{1}1+x^2dx


=> [1+\frac{1^3}{3}] = \frac{4}{3}

 

[OlderDan]

That looks good.