Integrate over area enclosed by curve

sanitykey

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]

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 [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

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

[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

That looks good.