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Math Help - Evaluate using spherical coordinates.

  1. #1
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    Evaluate using spherical coordinates.

    Evaluate using spherical coordinates: \int \int \int dx dy dz
    T: 0 \le x \le 1 \; \; ; \; \; 0 \le y \le \sqrt{1-x^2} \; \; ;\; \; \sqrt{x^2 + y^2} \le z \le \sqrt{2 - (x^2 + y^2)}

    I can represent this in cartesian coordinates like this:

    \int_0^1 \int_0^{\sqrt{1-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{2-(x^2+y^2)}}dzdydx

    and from here convert to spherical coordinates. I'm not sure how to completely convert this to spherical coordinates.

    The only thing I could see was that on the xy-plane, part of the area is a quarter of a circle in the first quadrant with radius 1 (since y goes from 0 to \sqrt{1-x^2} and x goes from 0 to 1), so \theta would go from 0 to \frac{\pi}{2}.
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Evaluate using spherical coordinates.

    drawing the area will help you how I draw it first I begin with the x-axis x changes from 0 to 1 as the drawing

    Evaluate using spherical coordinates.-first-x.jpg

    then we will draw y , y changes from 0 to sqrt{1-x^2} . sub x=0 and x=1 which will give us quarter of circle in xy plane in the positve for xy

    Evaluate using spherical coordinates.-first-xy.jpg

    now z the hardest one first let see what is sqrt{x^2+y^2} = z if x=0 , y=1 , z=1 and if y=0,x=1,z=1
    if z=1 this will give us a circle in xy plane with radius 1 but we interested just in xy positive

    Evaluate using spherical coordinates.-first-xyz.jpg

    now we will see z= sqrt{2-x^2-y^2 } , if x=0,y=0, then z=sqrt{2}
    it is sphere but positive z half of the sphere but since x,y are positive too so it is the quarter of the sphere
    z=sqrt{2-x^2-y^2} will intersect with z=sqrt{x^2+y^2} at z=1 the circle in xy plane

    Evaluate using spherical coordinates.-first-xyz1.jpg

    Evaluate using spherical coordinates.-first-xyx33x.jpg

    the area is the area with the purple color
    now can you determine the boundaries ?
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  3. #3
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    Re: Evaluate using spherical coordinates.

    z= \sqrt{x^2+ y^2} is the part of z^2= x^2+ y^2 above the xy-plane. As an aid to seeing what that is, note that when y= 0, it gives z^2= x^2 or z= \pm x and similarly for y. That is, this is a cone making an angle 0f \pi/4 with the z-axis. The other boundary, z= \sqrt{2- x^2- y^2} is the part of z^2= 2- x^2- y^2 or x^2+ y^2+ z^2= 2 above the xy-plane. That is, of course, a sphere of radius \sqrt{2}. Yes, \theta goes from 0 to \pi/2. \rho goes from 0 to \sqrt{2} and \phi from 0 to \pi/4.
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