# Integrate over area enclosed by curve

1. Nov 8, 2006

### 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 $$64\pi/3$$

Last edited: Nov 8, 2006
2. Nov 8, 2006

### 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.

3. Nov 8, 2006

### 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.

Last edited: Nov 8, 2006
4. Nov 8, 2006

### OlderDan

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.

Last edited: Nov 8, 2006
5. Nov 9, 2006

### sanitykey

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

6. Nov 9, 2006

### 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:

$$\frac{dy}{dx}=x$$

=> $$dy=xdx$$

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

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

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

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

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

=> $$\int1+x^2dx$$

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

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

Last edited: Nov 9, 2006
7. Nov 9, 2006

### OlderDan

That looks good.