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anonymous

  • 4 years ago

Can someone plz help me out with inequality using polar coordinates?

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  1. anonymous
    • 4 years ago
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  2. TuringTest
    • 4 years ago
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    um... not really sure if I get the question, but the area swept out by r is\[\frac38\le r\le\frac12\]and theta is just\[0\le\theta\le2\pi\]

  3. TuringTest
    • 4 years ago
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    no sorry inner r is 3/16

  4. TuringTest
    • 4 years ago
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    \[\frac3{16}\le r\le\frac12\]\[0\le\theta\le2\pi\]I suppose

  5. anonymous
    • 4 years ago
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    yes that is the answer

  6. TuringTest
    • 4 years ago
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    do you need explanation?

  7. anonymous
    • 4 years ago
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    lol yes i dont understand how u got thise values

  8. TuringTest
    • 4 years ago
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    well polar coordinates are defined with radius and angle. Here are the diameters|dw:1329346620272:dw|so we know the radii are half this. therefor radius of any point on the washer is between 3/16inches (the radius of the inner circle) and 1/2inches (the radius of the outer circle)\[\frac3{16}\le r\le\frac12\]

  9. anonymous
    • 4 years ago
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    ok got that

  10. TuringTest
    • 4 years ago
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    |dw:1329346791420:dw|as for the angle, the washer is circular, so the angle has to sweep out a complete circle:\[0\le\theta\le2\pi\](we never go around the circle more than once, so no need to put more than 2pi))

  11. anonymous
    • 4 years ago
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    ok i think i get it

  12. TuringTest
    • 4 years ago
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    if it was top half of a washer, the angle would be \[0\le\theta\le\pi\]if it was a disk (no hole in the middle) the radius would be\[0\le r\le\frac12\]

  13. anonymous
    • 4 years ago
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    ohhhh i see i get it

  14. anonymous
    • 4 years ago
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    oh it is so simple ok Thanks for helping me out :)

  15. TuringTest
    • 4 years ago
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    anytime!

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