# Integrate over area enclosed by curve

by sanitykey
Tags: curve, enclosed, integrate
 P: 93 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 $$64\pi/3$$
 Sci Advisor HW Helper 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.
 P: 93 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 :)
HW Helper
P: 3,033

## Integrate over area enclosed by curve

 Quote by 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 :)
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.
 P: 93 Thanks again my main problem with these types of questions is visualising what it actually means.
 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 $$\int\sqrt{1+x^2}ds$$ 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: $$\frac{dy}{dx}=x$$ => $$dy=xdx$$ => $$dy^2 = x^2 dx^2$$ $$\int\sqrt{1+x^2}\sqrt{dx^2 + dy^2}$$ => $$\int\sqrt{1+x^2}\sqrt{dx^2 + x^2 dx^2}$$ => $$\int\sqrt{1+x^2}\sqrt{dx^2 (1+x^2)}$$ => $$\int\sqrt{1+x^2}\sqrt{1+x^2}dx$$ => $$\int1+x^2dx$$ => $$\int_{0}^{1}1+x^2dx$$ => $$[1+\frac{1^3}{3}] = \frac{4}{3}$$
 Sci Advisor HW Helper P: 3,033 That looks good.

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