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Double integration of circles, spheres etc

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data
    So I have three questions. The first one is the double integration of x+2 from y=0 to y=sqrt(9^2-x^2). The second question is the double integration of sqrt(r^2-x^2-y^2) where the domain is in the circle of radius R and origin 0. And the last question is the double integration of ax^3+by^3+sqrt(a^2-x^2), where xis from -a to a and y is from -b to b. Please no polar coordinates!


    2. Relevant equations

    3. The attempt at a solution
    1. I drew a semicircle where the radius was 3. The x is therefore from -3 to 3. And the y can be simplified to be from 0 to 3*pi, since the outer border of the semicircle is half the circumference, which is 2*pi*r. So, (2*pi*3)/2 is 3*pi. Therefore x is from -3 to 3 and y is from 0 to 3*pi. After I integrated, first with respect to y, then to x, I got 36*pi. But the textbook answer is 9*pi. Why?

    2. I don't really know where to start. I just know that I'll have z^2+x^2+y^2=r^2 and that on the xy plane z=0 so I have x^2+y^2=r^2 which is one of the boundaries. So I can have the boundaries 0≤y≤sqrt(r^2-x^2) and -r≤x≤r, say if I'm only focusing on the semicircle which lies on the xy plane. But after that I don't know what to do!

    3. I first simplified the formula by noticing the sqrt part can be drawn as a semicircle, with radius a. This means that the circumference of that semicircle will be 2*pi. So I can put in 2*pi and get rid of the sqrt equation. Then by integrating first with respect to y, then to x, and using the boundaries I mentioned, I get 4*a^2*b*pi. I don't know how to get rid of the 4!

    Thanks.
     
  2. jcsd
  3. Nov 2, 2012 #2
    This is where you go wrong. You can't have fixed limits for y here, because it is from 0 to the border of the circle, and that depends on x. You need to integrate w.r.t. y first, which is trivial because the integrand does not depend on it, then plug in the y-limits, which are functions of x, then integrate by x.

    The integral over the entire circle is the sum of the integrals over its semicircles. And you already know how to tackle those.

    No, this is completely wrong. The area of integration is a rectangle, not a circle (unless you change coordinates). The area makes things simple in the sense that the limits of integration are constant, so the double integral becomes trivial. Just integrate w.r.t. one variable first, assuming the other is constant, then integrate by the other one.
     
  4. Nov 2, 2012 #3
    No sorry I don't understand. Could you specifically say what boundaries I would have to use? I don't see how I'd use the y-limits for the first question as you said, unless I use polar coordinates which I'm not supposed to do, because this question is part of a set with rectangular coordinates only.
     
  5. Nov 3, 2012 #4
    For the first case, you have this: [tex]

    \int_{x = -3}^{x = 3} dx \int_{y = 0}^{y = \sqrt{9-x^2}} (x + 2)dy

    [/tex] You first integrate with respect to y: [tex]

    \int_{x = -3}^{x = 3} dx \int_{y = 0}^{y = \sqrt{9-x^2}} (x + 2)dy

    = \int_{x = -3}^{x = 3} dx \left[ (x + 2)y \right]_{y = 0}^{y = \sqrt{9-x^2}}

    = \int_{x = -3}^{x = 3} dx (x + 2)\sqrt{9-x^2}

    [/tex] This is on ordinary definite integral.
     
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