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Math Help - Converting Cartersian coordinate term to spherical

  1. #1
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    Converting Cartersian coordinate term to spherical

    I am working through a problem where the vector has a Cartesian term and want to convert it to spherical::
    (Sorry, I couldn't figure out how to use the equation editor)

    sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) (x component of vector)

    I know:
    r=sqrt(x^2+y^2+z^2)
    x=rsin\theta cos\phi
    y=rsin\theta sin\phi
    z=rcos\theta

    I also know the answer is:
    (rsin\phi )/r = sin\phi

    I can't figure out how rsin\phi = sqrt(x^2=y^2) in the answer.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by laguna92651 View Post
    I am working through a problem where the vector has a Cartesian term and want to convert it to spherical::
    (Sorry, I couldn't figure out how to use the equation editor)

    sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) (x component of vector)

    I know:
    r=sqrt(x^2+y^2+z^2)
    x=rsin\theta cos\phi
    y=rsin\theta sin\phi
    z=rcos\theta

    I also know the answer is:
    (rsin\phi )/r = sin\phi

    I can't figure out how rsin\phi = sqrt(x^2=y^2) in the answer.

    How this is possible that the first red is NOT an equation and the second red is an equation?
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  3. #3
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    The second red, just goes one step further to simplify the answer.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by laguna92651 View Post
    The second red, just goes one step further to simplify the answer.
    Let me put my question in other words, where is the "=" sign in the first red?
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  5. #5
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    I'll ask the question is way, how did the left convert to the right side? Cartesian to spherical

    sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) = [r(sin(phi) ]/r
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  6. #6
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    Quote Originally Posted by laguna92651 View Post
    I'll ask the question is way, how did the left convert to the right side? Cartesian to spherical

    sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) = [r(sin(phi) ]/r
    sqrt(x^2+ y^2+ z^2)= r. that's where the denominator on the right comes from.

    x= r cos(theta) sin(phi), y= r sin(theta) sin(phi) so x^2= r^2 cos^2(\theta) sin^2(phi) and y^2= r^2 sin^2(\theta) sin^2(\phi).

    x^2+ y^2= r^2 cos^2(theta) sin^2(phi)+ r^2 sin^2(theta) cos^2(phi)= r^2 sin^2(phi)(cos^2(theta)+ sin^2(theta))
    = r^2 sin^2(\phi). That's the numerator.
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  7. #7
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    Thanks for the help.

    Did you mean for the second term in the x^2+y^2 equation to be sin^2(phi) instead of cos^2(phi)?
    Second term: r^2 sin^2(theta) cos^2(phi)

    You used; x= r cos(theta) sin(phi)
    My electromagnetics book uses; x= r cos(phi) sin(theta)

    So I get sin(theta) as an answer, the correct answer is sin(phi). Am I missing something?

    Theta is defined as the angle between the z-axis the position vector.
    Phi is measured from the x-axis to the plane of the vector.
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