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blackstreet23

  • one year ago

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

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  1. blackstreet23
    • one year ago
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  2. TheCatMan
    • one year ago
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    im not that advanced in school yet i cant help here sorry

  3. TheCatMan
    • one year ago
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    @whpalmer4 can you help

  4. anonymous
    • one year ago
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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?

  5. blackstreet23
    • one year ago
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    i dont get it

  6. blackstreet23
    • one year ago
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    why pi/4?

  7. blackstreet23
    • one year ago
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    @SithsAndGiggles

  8. blackstreet23
    • one year ago
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    @Shalante

  9. blackstreet23
    • one year ago
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    @pooja195

  10. anonymous
    • one year ago
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    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.

  11. blackstreet23
    • one year ago
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    Thanks a lot !!! you are awesome :D

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