Write down, but do not evaluate, an integral for the area of this shaded region.

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Write down, but do not evaluate, an integral for the area of this shaded region.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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im not that advanced in school yet i cant help here sorry
@whpalmer4 can you help

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|dw:1442977155087:dw| This means the upper half of this region can be given by the double integral \[\int_0^{\pi/4}\int_0^{\cos2 t}r\,dr\,dt=\frac{1}{2}\int_0^{\pi/4}\cos^2t\,dt\] How might you adjust the limits to account for the bottom half as well?
i dont get it
why pi/4?
Your curve is defined by the function \(r(t)=\cos2t\). When \(t=\dfrac{\pi}{4}\), you have \[r\left(\frac{\pi}{4}\right)=\cos\frac{2\pi}{4}=\cos\frac{\pi}{2}=0\] (You can check that \(r(t)\neq0\) for values of \(t\) between \(0\) and \(t=\dfrac{\pi}{4}\).)|dw:1443051824798:dw| As \(t\) increases from \(0\) to \(\dfrac{\pi}{4}\), the curve is traced out in the direction indicated by the arrows.
Thanks a lot !!! you are awesome :D

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