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Math Help - Method for converting integral:cylindrical coord to rectangular coord

  1. #1
    Senior Member x3bnm's Avatar
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    Method for converting integral:cylindrical coord to rectangular coord

    I want to convert integrals \int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta from cylindrical coordinate
    to rectangular coordinate( x,y and  z). Is there an easy way to convert these integrals from cylindrical
    coordinate to rectangular coordinate?

    What is the best method for this conversion?

    The answer for the above problem is: \int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx + \int_1^{\sqrt{3}} \int_0^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx

    Thanks.
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  2. #2
    Senior Member x3bnm's Avatar
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    For converting
    \int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta
    integral from cylindrical coordinate to rectangular coordinate first i find the limits(upper/lower boundaries of x,y and z).

    We know that (from the integral)
    <br />
  z = \sqrt{4-r^2}<br />
    so we can say the upper limit of z is:
    <br />
 z = \sqrt{4-x^2-y^2}<br />
    and lower limit of z is:
    <br />
 z = 1 <br />


    Also we know from the integral that
    <br />
 r= \sqrt{3}<br />
    So we can say:
     <br />
\sqrt{(r \sin(\theta))^2 +(r \cos(\theta))^2} = \sqrt{3}<br />
    And:
    <br />
 y^2 + x^2 = 3<br />
    As a result the upper limit of y is:
    <br />
y = \sqrt{3 - x^2}<br />
    Again from the integral r = 1
    Using the same procedure we can say lower limit of y is: y = \sqrt{1 - x^2}


    Upper limit of x is
    <br />
x  = \sqrt{3}<br />
    and lower limit of x is
    <br />
 x = 1<br />


    Now my result is:
    <br />
\int_1^{\sqrt{3}} \int_{\sqrt{1 - x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4-x^2-y^2}}  \,\, z^2yx\,\, dz\,\, dy\,\, dx<br />


    But the answer is different. It is:
    \int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx + \int_1^{\sqrt{3}} \int_0^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx<br />

    Why the answer is different? Can anyone kindly point out where i did wrong?
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  3. #3
    Senior Member x3bnm's Avatar
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    If I convert \int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta

    to rectangular coordinate can it be equal to

    \int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx + \int_1^{\sqrt{3}} \int_0^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx<br />
 ?

    Please a simple yes or no will do.
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  4. #4
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    z= \sqrt{4- r^2} is almost the same as z^2= 4- r^2= 1- x^2- y^2 or x^2+ y^2+ z^2= 4, the sphere with center at (0, 0, 0) and radius 2. "Almost" because, of course, the square root is positive so this is actually the hemi-sphere above the xy-plane. The lower limit on z is just z= 1.

    Taking \theta from 0 to \pi/2 and r from 1 to \sqrt{3} means that we, in the x, y plane, we are integrating between the circles x^2+ y^2= 1 and x^2+ y^2= 3 in the first quadrant.

    If we want to do this integral in the order \int\int\int dzdydx we will need to break it into two parts. For x between 0 and 1, the lower boundary is the smaller circle, x^2+ y^2= 1 or y= \sqrt{1- x^2}, since we are in the first quadrant, and the upper boundary is the larger circle, x^2+ y^2= 3 or y= \sqrt{3- x^2} since we are in the first quadrant. For x between 1 and \sqrt{3}, the upper boundary is still y= \sqrt{3- x^2} but the lower boundary is y= 0. In either integral, z goes from 1 to \sqrt{4- x^2- y^2}.

    The limits of integration will be \int_{x=0}^1\int_{y= \sqrt{1- x^2}}^{\sqrt{3- x^2}}\int_{z=1}^{\sqrt{4- x^2- y^2}} dz dy dx+ \int_{x= 1}^{\sqrt{3}}\int_{y= 0}^{\sqrt{3- x^2}}\int_{z= 1}^{\sqrt{4- x^2- y^2}} dz dy dx.

    Your integrand is r^3 sin(\theta)cos(\theta)z^2 dz dr d\theta but the differential of volume in cylindrical coordinates is r dz dr d\theta so we can write that as r^2 sin(\theta)cos(\theta)z^2 (r dz dr d\theta)= (r sin(\theta))(r cos(\theta)) z^2 dz dy zr= xyz^2 dz dy dx

    Putting that all together,
    \int_{x=0}^1\int_{y= \sqrt{1- x^2}}^{\sqrt{3- x^2}}\int_{z=1}^{\sqrt{4- x^2- y^2}} xyz^2 dz dy dx+ \int_{x= 1}^{\sqrt{3}}\int_{y= 0}^{\sqrt{3- x^2}}\int_{z= 1}^{\sqrt{4- x^2- y^2}} xyz^2 dz dy dx

    which is just what you have! Well done!
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  5. #5
    Senior Member x3bnm's Avatar
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    HallsofIvy, all i needed is your simple explanation. And thank you for that.
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