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Math Help - Triple Integral

  1. #1
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    Triple Integral

    Let A be the region that in space bounded by the balls:
     x^2 +y^2 + z^2 =1 ,  x^2 +y^2 +z^2 =4 , above the plane z=0 and inside the cone z^2 = x^2 +y^2 .
    A. Write the integral  \int \int \int_{A} f(x,y,z) dxdydz in the form:
     \int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z) dz) dxdy when :
     A=( (x,y,z) | (x,y) \in E, g^1(x,y) \le z \le g^2(x,y) ) ...
    B. Find the volume of A (not necessarily using part A).
    Hope you'll be able to help me in this... I think the main problem is that I can't figure out how A looks like... There is also a hint that one of the functions g1 or g2 should be defined at a split region... I can't figure out how this cone looks like and how I can describe A as equations ...

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    Let A be the region that in space bounded by the balls:
     x^2 +y^2 + z^2 =1 ,  x^2 +y^2 +z^2 =4 , above the plane z=0 and inside the cone z^2 = x^2 +y^2 .
    A. Write the integral  \int \int \int_{A} f(x,y,z) dxdydz in the form:
     \int \int_{E} (\int_{g^1(x,y)}^{g^2(x,y)} f(x,y,z) dz) dxdy when :
     A=( (x,y,z) | (x,y) \in E, g^1(x,y) \le z \le g^2(x,y) ) ...
    B. Find the volume of A (not necessarily using part A).
    Hope you'll be able to help me in this... I think the main problem is that I can't figure out how A looks like... There is also a hint that one of the functions g1 or g2 should be defined at a split region... I can't figure out how this cone looks like and how I can describe A as equations ...

    Thanks in advance!
    The cone intersects the inner sphere, x^2+ y^2+ z^2= 1 where x^2+ y^2+  (x^2+ y^2)= 2(x^2+ y^2)= 1, the circle of radius \sqrt{2}/2, in the plane z= 1/2, with center at (0, 0, 1/2). It intersects the outer sphere, x^2+ y^2+ z^2= 4 where x^2+ y^2+ (x^2+ y^2)= 2(x^2+ y^2)= 4, the circle of radius \sqrt{2}, in the plane z= 2, with center at (0, 0, 2).

    The volume inside the cylinder, x^2+ y^2= 1/2, is easy:
    \int_{x= -\sqrt{2}/2}^{\sqrt{2}/2}\int_{y= -\sqrt{1/2- x^2}}^{\sqrt{1/2- x^2}}\int_{z=\sqrt{1- x^2- y^2}}^{\sqrt{1- x^2- y^2}} dzdydx.

    But that misses the volume, above the cone, that lies in the annulus between the two circles, x^2+ y^2= 1 and x^2+ y^2+ z^2= 4. To do that you will need to separate it into four integrals:

    The first with x from -2 to -\sqrt{2}/2, y from -\sqrt{4- x^2} to \sqrt{4- x^2}, z from \sqrt{x^2+ y^2} to \sqrt{4- x^2- y^2}

    The second with x from -\sqrt{2}/2 to \sqrt{2}/2, y from \sqrt{1- x^2} to \sqrt{4- x^2}, z from \sqrt{x^2+ y^2} to \sqrt{4- x^2- y^2}

    The third with x from -\sqrt{2}/2 to \sqrt{2}/2, y from -\sqrt{4- x^2} to -\sqrt{1- x^2},z from \sqrt{x^2+ y^2} to \sqrt{4- x^2- y^2}

    The fourth with x from \sqrt{2}/2 to \sqrt{2}, y from -\sqrt{4- x^2} to \sqrt{4- x^2},z from \sqrt{x^2+ y^2} to \sqrt{4- x^2- y^2}

    That might make more sense to you if you draw the circles in the xy-plane and mark out each of the five regions above

    To actually find the volume, it would be much easier to use spherical coordinates. In spherical coordinates, \rho goes from 1 to 2, \theta goes from 0 to 2\pi, and \phi goes from 0 to 2\pi.
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