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Double integral to find area

  1. Dec 4, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-12-4_6-20-17.png

    2. Relevant equations


    3. The attempt at a solution
    here is my approach,

    I take the whole area, which is π16

    then subtract the unshaded region

    now to find the unshaded region's area, I use rectangular coordinates.

    my bounds are from -2 to 2 for x and the the top and bottom of the larger circle : sqrt (16-x^2) and -sqrt (16-x^2)

    but now I don't know if there should be an integrand or assume it is 1 if none is given?
     
  2. jcsd
  3. Dec 4, 2015 #2

    RUber

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    When you integrate for area, the integrand is 1.
     
  4. Dec 6, 2015 #3
    Here's how I like to think of it. If the amount of integrals that you're doing is the same as the exponent on your units would be, then the integrand is 1. For example, if you're calculating an area, your units would be square units (exponent 2), so with a double integral the integrand is one. So if you're calculating a volume, with a triple integral your integrand is 1 but with a double integral it isn't.
     
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