Question #e4aa5

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Question #e4aa5

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Answer 1 · 2017-01-24T22:12:37

A point $P$ $P$ is defined in spherical coordinated as a function of $(r, θ, ϕ)$ $(r, theta, phi)$ . Where $r$ $r$ is the distance from the origin to the point, $ϕ$ $phi$ is the angle that we need to rotate down from the positive $z$ $z$ -axis to get to the point and $θ$ $theta$ is angle we need to rotate around the $z$ $z$ -axis to get to the point.

As you have rightly posted the figure, infinitesimal volume element $d V$ $dV$ in spherical coordinates is given by multiplication of three unit coordinates

Infinitesimal radial element $= d r$ $=dr$
Recall formula which relates the arc length $s$ $s$ of a circle of radius $r$ $r$ to the central angle $θ$ $theta$ is given by
$s = r θ$ $s=rtheta$ , where angle is in radians.
Hence, arc length subtended by angle $d θ$ $d theta$ at a distance $r$ $r$ would be $= r d θ$ $=rd theta$
Consider area element as shown in the figure below

We see that area element is located at a perpendicular distance $= r sin θ$ $=r sin theta$ from the $z$ $z$ -axis.
Using the same formula as used in step 2. above we see that arc length subtended by angle $d ϕ$ $d phi$ at a distance of $r sin θ$ $rsin theta$ would be $= r sin θ d ϕ$ $=rsin theta dphi$
Therefore we get

$d V = d r \cdot r d θ \cdot r sin θ d ϕ$ $dV=drcdot rd theta cdot rsinthetadphi$
$\Rightarrow d V = r^{2} sin θ \cdot d r \cdot d θ \cdot d ϕ$ $=>dV=r^2 sin theta cdot dr cdot d theta cdot dphi$

$r$ $r$ ranges from $0$ $0$ to $r$ $r$
$θ$ $theta$ from $0$ $0$ to $π$ $pi$ and
$ϕ$ $phi$ from $0$ $0$ to $2 π$ $2pi$

A word of caution : Sometimes $∠ θ and ∠ ϕ$ $angle theta and angle phi$ are interchanged in Physics and Mathematics.

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