(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the area integral of the surface [tex]z=y^2+2xy-x^2+2[/tex] in polar form lying over the annulus [tex]\frac{3}{8}\leq x^2+y^2\leq1[/tex]

2. Relevant equations

The equation in polar form is [tex]r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2[/tex]

[tex]\displaystyle{\int^{\pi}_{-\pi}}\int^1_{\sqrt{\frac{3}{8}}}r^2\sin^2\theta+2r^2\cos\theta\sin\theta-r^2\cos^2\theta+2\, r\, dr\, d\theta[/tex]

3. The attempt at a solution

Hi, would just like to know if what I have done is correct.

Integrating with respect to [tex]r[/tex]

[tex]{\frac {55}{256}}\, \left( \sin \left( \theta \right) \right) ^{2}+{

\frac {55}{128}}\,\cos \left( \theta \right) \sin \left( \theta

\right) -{\frac {55}{256}}\, \left( \cos \left( \theta \right)

\right) ^{2}+5/8

[/tex]

and with respect to [tex]\theta[/tex], I get the area as

[tex]\frac{5}{4}\pi[/tex]

Does this look ok?

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# Homework Help: Surface integral over an annulus

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