anonymous
  • anonymous
Use a double integral to find the area of the region that has the indicated shape. One loop of r = 2cos(4theta)
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
we need to find a smallest interval where cos 4t goes form 0 to 0
amistre64
  • amistre64
the other integral is just measureing r from 0 to the function itself
amistre64
  • amistre64
\[\text{when t=pi/2 and 3pi/2; cos t = 0}\] \[\frac{4}{4}\frac{pi}{2}=4\frac{pi}{8}\] \[\frac{4}{4}\frac{3pi}{2}=4\frac{3pi}{8}\] so the smallest interval i see is from pi/8 to 3pi/8

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amistre64
  • amistre64
\[\large \int_{t=pi/8}^{t=3pi/8}\int_{r=0}^{r=2cos(4t)}r\ dr.dt\]
amistre64
  • amistre64
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