- #1

sanitykey

- 93

- 0

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]

[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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