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Thread: 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 $\displaystyle \int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta$ from cylindrical coordinate
    to rectangular coordinate($\displaystyle x,y$ and $\displaystyle z$). 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: $\displaystyle \int_0^1 \int_{\sqrt{1-x^2}}^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx + \int_1^{\sqrt{3}} \int_0^{\sqrt{3-x^2}} \int_1^{\sqrt{4 - x^2 -y^2}}\,\, z^2yx\,\, dz\,\, dy\,\, dx$

    Thanks.
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  2. #2
    Senior Member x3bnm's Avatar
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    For converting
    $\displaystyle \int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta$
    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)
    $\displaystyle
    z = \sqrt{4-r^2}
    $
    so we can say the upper limit of $\displaystyle z$ is:
    $\displaystyle
    z = \sqrt{4-x^2-y^2}
    $
    and lower limit of $\displaystyle z$ is:
    $\displaystyle
    z = 1
    $


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


    Upper limit of $\displaystyle x$ is
    $\displaystyle
    x = \sqrt{3}
    $
    and lower limit of $\displaystyle x$ is
    $\displaystyle
    x = 1
    $


    Now my result is:
    $\displaystyle
    \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
    $


    But the answer is different. It is:
    $\displaystyle \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
    $

    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 $\displaystyle \int_0^{\frac{\pi}{2}} \int_1^{\sqrt{3}} \int_1^{\sqrt{4 - r^2}}\,\, r^3(\sin(\theta) \cos(\theta))z^2\,\, dz\,\, dr\,\, d\theta$

    to rectangular coordinate can it be equal to

    $\displaystyle \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
    $ ?

    Please a simple yes or no will do.
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  4. #4
    MHF Contributor

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    $\displaystyle z= \sqrt{4- r^2}$ is almost the same as $\displaystyle z^2= 4- r^2= 1- x^2- y^2$ or $\displaystyle x^2+ y^2+ z^2= 4$, 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 $\displaystyle \theta$ from 0 to $\displaystyle \pi/2$ and r from 1 to $\displaystyle \sqrt{3}$ means that we, in the x, y plane, we are integrating between the circles $\displaystyle x^2+ y^2= 1$ and $\displaystyle x^2+ y^2= 3$ in the first quadrant.

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

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

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

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

    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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