- #1
bobred
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Homework Statement
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]
Homework 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]
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?