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Thread: Converting Coordinate Systems

  1. #1
    Super Member craig's Avatar
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    Converting Coordinate Systems

    Bit stuck on converting from Cartesian to Spherical:

    Let $\displaystyle S$ be the spherical surface defined by $\displaystyle x^2 + y^2 + z^2 = a^2$, and let $\displaystyle \mathbf{F} = z^3 \mathbf{k}$.

    Using SPs $\displaystyle (r,\theta,\phi)$, and with $\displaystyle \mathbf{\delta S} = a^2 \sin{\theta} \delta \theta \delta \phi \mathbf{\hat{r}}$

    i) Show $\displaystyle \mathbf{F} \cdot \mathbf{\delta S} = a^5 \cos^4{\theta} \sin{\theta} \delta \theta \delta \phi$

    ii) Calculate $\displaystyle \int_S \mathbf{F} \cdot \mathbf{dS}$.

    i)Well converting $\displaystyle \mathbf{F}$ to Spherical, we get $\displaystyle (z^3,0,0)$, so the scalar product gives:

    $\displaystyle z^3(a^2 \sin{\theta} \delta \theta \delta \phi)$.

    Now we also know that $\displaystyle z = a \cos{\theta}$, either from using $\displaystyle z = a^2 - (x^2 + y^2)$, or from the definition of SPs.

    So substituting this in gives:

    $\displaystyle \cos^3{\theta} a^3(a^2 \sin{\theta} \delta \theta \delta \phi)$

    Which simplifies to:

    $\displaystyle a^5 \cos^3{\theta} \sin{\theta} \delta \theta \delta \phi$

    I'm just at a bit of a loss where they get their extra $\displaystyle \cos{\theta}$ from?

    ii) Integrating this gives us $\displaystyle a^5 \int^{2 \pi}_0 d\phi \int^{pi}_0 \cos^4{\theta} \sin{\theta} d\theta$

    $\displaystyle \frac{-2 \pi a^5}{5} \left[\cos^5{\theta} \right]^{\pi}_0$

    Which gives $\displaystyle \frac{4 \pi a^5}{5}$, would this be right?

    Thanks in advance for any help
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  2. #2
    Super Member craig's Avatar
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    Would the first bit involving the extra $\displaystyle \cos{\theta}$ be because:

    $\displaystyle \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos{\theta}$?

    I think this would explain where they've got the extra $\displaystyle \cos{\theta}$ from?
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