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blackstreet23
 one year ago
Write down, but do not evaluate, an integral for the area of this shaded region.
blackstreet23
 one year ago
Write down, but do not evaluate, an integral for the area of this shaded region.

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TheCatMan
 one year ago
Best ResponseYou've already chosen the best response.0im not that advanced in school yet i cant help here sorry

TheCatMan
 one year ago
Best ResponseYou've already chosen the best response.0@whpalmer4 can you help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw: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?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your 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.

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0Thanks a lot !!! you are awesome :D
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