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Area is symmetric

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Help!! I was wondering if anyone can help me integrate:

    ∫ x^2tan x + y^3 +4 dA, where D is the region represented by D = {(x,y)|x2+y2≤2}

    2. Relevant equations
    I think that the area is symmetric, and so basically you only need to evalute from 0≤ x ≤ √(2-y^2) and 0≤ y ≤ √2. Or you can do x first and evaluate it from 0 ≤ y ≤ √ (1-x^2) and 0≤ x ≤ √2. But I'm not sure how to evaluate x^2 tan x? I don't think doing dy first will help either since I get the following: x^2*√ (1-x^2)*tan x? Thanks!!!
  2. jcsd
  3. Jan 30, 2007 #2

    Tom Mattson

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    I've been playing around with this, and I think I've made some progress.

    Integrate the first term in the integrand with respect to [itex]x[/itex] first. Do it by parts with [itex]u=x \tan(x)[/itex] and [itex]dv=x dx[/itex]. You should be able to express the integral in terms of [itex]\int x^3\sec^2(x)dx[/itex]. Integrate that by parts with [itex]u=x^3[/itex] and [itex]dv=sec^2(x)dx[/itex].

    Give that a try and see how it goes. I haven't finished it yet, but it looks like it will work.
  4. Jan 31, 2007 #3


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    This is, of course, the same as
    [tex]\int x^2 tan x dA+ \int y^3 dA+ \int 4 dA[/tex]
    Yes, the region [itex]D= {(x,y)| x^2+ y^2\le 2}[/itex] is a circle and so is symmetric. But why do you then say you need only integrate in the first quadrant? x2tan x and y3 are both ODD functions. Their integrals on opposite sides of the axes will cancel, not add. It looks to me like this is just [itex]\int 4 dA[/itex] or just 4 times the area of the circle.
  5. Jan 31, 2007 #4
    Oh I see!! Thanks!!!
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