Helpful identities
This is a collection of identities, proofs, etc. to aid the presentation in the previous sections.
Standard vector identities
(1)\[ \renewcommand{\arraystretch}{2} \begin{array}{rcl}
{\bf \nabla} \left( {\bf a}\cdot {\bf b} \right) = \left( {\bf a}\cdot {\bf \nabla} \right) {\bf b}
+ \left( {\bf b}\cdot {\bf \nabla} \right) {\bf a}
+ {\bf a} \times \left( {\bf \nabla} \times {\bf b} \right)
+ {\bf b} \times \left( {\bf \nabla} \times {\bf a} \right)
\end{array}\]
Vector identities
Identity 1
For a scalar field \(a({\bf y})\) with
\({\bf y} = y_1{\bf\hat{x}_1} + y_2{\bf\hat{x}_2} + z{\bf\hat{x}_3}\)
a vector from the origin
to point \((y_1,y_2,y_3)\),
if \(F \equiv \left( \nabla^2 a({\bf y}) \right) {\bf y}\) then if follows that
:math:` F = left( sum_i partial_i^2 a({bf y}) right) {bf y}mbox{ }` and
(2)\[\begin{split} \renewcommand{\arraystretch}{1} \begin{array}{rcl}
F_j & = & \sum_i \partial_i^2 a({\bf y}) \; y_j \\
F_j & = & \sum_i \partial_i \left(\partial_i a({\bf y}) \; y_j \right)
- \sum_i \partial_i a({\bf y}) \; \partial_i y_j \\
F_j & = & \sum_i \partial_i \left(\partial_i a({\bf y}) \; x_j \right)
- \partial_j a({\bf x}) y_j,
\end{array}\end{split}\]
since \(\partial_i y_j=\delta_{ij} y_j\).
But
(3)\[\begin{split} \renewcommand{\arraystretch}{1} \begin{array}{rcl}
F_j & = & \sum_i \partial_i \left(\partial_i a({\bf x}) \; y_j \right)
- \partial_j a({\bf y}) y_j
= \sum_i \partial_i \left(y_j \; \partial_i a({\bf y}) \right)
- \partial_j a({\bf y}) y_j
\end{array}\end{split}\]
Then
(4)\[ \boxed{ \renewcommand{\arraystretch}{1} \begin{array}{rcl}
\sum_i \partial_i \left(\partial_i a({\bf y})\; {\bf y} \right)
= \sum_i \partial_i \left(({\bf y} \; \partial_i a({\bf y}) \right)
= \left( \nabla^2 a({\bf y}) \right) {\bf y} + \nabla a \; {\bf y}
\end{array} }\]
Divergence theorem,
(5)\[\begin{split} \renewcommand{\arraystretch}{1} \begin{array}{rcl}
\int_V dV\, \left( {\bf \nabla} \cdot {\bf A} \right)
& = & \int_S dS\, {\bf A} \cdot {\bf \hat{n}},
\end{array}\end{split}\]
where \({\bf A}\) is a vector and \({\bf \hat{n}}\) is the normal to the surface.
Identity 2
Ball \(B\) of radius \(\rho\) centered at the origin,
Scalar field \(a( {\bf y} )\) with
\({\bf y} = y_1{\bf\hat{x}_1} + y_2{\bf\hat{x}_2} + z{\bf\hat{x}_3}\)
a vector from the origin
to point \((y_1,y_2,y_3)\),
\({\bf x}=\rho {\bf \hat{x}}\).
(6)\[\begin{split} \boxed{ \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\int_V dV\; \left( \nabla^2 a({\bf y}) \right) {\bf y}
& = & \rho \int_S dS\; \left( {\bf \hat{x}} {\bf \hat{x}} \cdot {\bf \nabla} a(\rho {\bf \hat{x}} ) \right)
- \int_V dV\; \nabla a({\bf y}),
\end{array} }\end{split}\]
Coordinate and other transformations
Spherical coordinate convention:
- radial \(r\)
- \(\phi\) is the azimuthal angle - the angle in the \(xy\)-plane (\(0 \le \theta \le 2\pi\))
- \(\theta\) is the polar (zenith) angle - the angle with respect to the \(z\)-axis (\(-\pi \le \phi \le \pi\))
This is the convention commonly used in physics (and Wikipedia!),
while the mathematics conventions swap \(\theta\) and \(\phi\).
(http://mathworld.wolfram.com/SphericalCoordinates.html)
Unit vector identities
http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
(8)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\hat{r} & = & \frac{x\hat{x} + y\hat{y} + z\hat{z}}{\sqrt{x^2+y^2+z^2}} \\
\hat{\theta} & = & \frac{z(x\hat{x}+y\hat{y})-(x^2+y^2)\hat{z}}{\sqrt{x^2+y^2+z^2} \sqrt{x^2+y^2}} \\
\hat{\phi} & = & \frac{-y\hat{x}+x\hat{y}}{\sqrt{x^2+y^2}}
\end{array}\end{split}\]
(9)\[\begin{split} \renewcommand{\arraystretch}{1} \begin{array}{rcl}
\hat{x} & = & \cos\phi \sin\theta \,\hat{r} + \cos\phi \cos\theta \,\hat{\theta} - \sin\phi \,\hat{\phi} \\
\hat{y} & = & \sin\phi \sin\theta \,\hat{r} + \sin\phi \cos\theta \,\hat{\theta} + \cos\phi \,\hat{\phi} \\
\hat{z} & = & \cos\theta \,\hat{r} - \sin\theta \,\hat{\theta}
\end{array}\end{split}\]
Derivatives
http://mathworld.wolfram.com/SphericalCoordinates.html
(10)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\frac{\partial}{\partial x} & = & \cos\theta\, \sin\phi\, \frac{\partial}{\partial r}
- \frac{\sin\theta}{r\sin\phi} \frac{\partial}{\partial \theta}
+ \frac{\cos\theta\, \cos\phi}{r} \frac{\partial}{\partial \phi}\\
\frac{\partial}{\partial y} & = & \sin\theta\, \sin\phi\, \frac{\partial}{\partial r}
- \frac{\cos\theta}{r\sin\phi} \frac{\partial}{\partial \theta}
+ \frac{\sin\theta\, \cos\phi}{r} \frac{\partial}{\partial \phi}\\
\frac{\partial}{\partial z} & = & \cos\phi\, \frac{\partial}{\partial r}
+ \frac{\sin\phi}{r} \frac{\partial}{\partial \phi}
\end{array}\end{split}\]
(11)\[ \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\nabla f = \frac{\partial f}{\partial r} \hat{r}
+ \frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\theta}
+ \frac{1}{r\sin\theta} \frac{\partial f}{\partial \phi} \hat{\phi}
\end{array}\]
The (cartesian coordinates) Laplacian converted to spherical coordinates is
http://skisickness.com/2009/11/20/
(12)\[ \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\nabla^2 f = \frac{1}{r^2} \frac{\partial }{\partial r}\left( r^2 \frac{\partial f}{\partial r} \right)
+\frac{1}{r^2\sin\theta} \frac{\partial }{\partial \theta} \left( \sin\theta \frac{\partial f}{\partial \theta} \right)
+ \frac{1}{r^2\sin\theta} \frac{\partial^2 f}{\partial \phi^2} .
\end{array}\]
Identities involving the unit vector to the sphere
Consider a sphere of radius \(\rho\) centered at the origin,
a point \((x,y,z)\) on the sphere surface, \(x^2+y^2+z^2=\rho^2\), and
\({\bf \hat{u}} = \frac{ x \hat{x} + y \hat{y} + z \hat{z} }{ \left( x^2 + y^2 + z^2 \right)^{\frac{1}{2}} }\)
a unit vector from the origin.
Unit vector derivatives
(13)\[\begin{split} \renewcommand{\arraystretch}{2.0} \begin{array}{rcl}
\frac{\partial }{\partial x } \, {\bf \hat{u}} & = &
\frac{ - x \left( x \hat{x} + y \hat{y} + z \hat{z} \right) }{ \left( x^2 + y^2 + z^2 \right)^{\frac{3}{2}} }
+ \frac{\hat{x} }{ \left( x^2 + y^2 + z^2 \right)^{\frac{1}{2}} }
= \frac{1}{\rho} \left( - u_x {\bf \hat{u}} + \hat{x} \right) \\
\frac{\partial }{\partial y } \, {\bf \hat{u}} & = & \frac{1}{\rho} \left( - u_y {\bf \hat{u}} + \hat{y} \right) \\
\frac{\partial }{\partial z } \, {\bf \hat{u}} & = & \frac{1}{\rho} \left( - u_z {\bf \hat{u}}+ \hat{z} \right)
\end{array}\end{split}\]
Curl of unit vector to sphere surface
Given the gradient operator
\(\nabla f = \frac{\partial f}{\partial x} \hat{x} + \frac{\partial f}{\partial y} \hat{y} + \frac{\partial f}{\partial z} \hat{z}\), show that
(14)\[ \boxed{ \renewcommand{\arraystretch}{1.0} \begin{array}{rcl}
{\bf \nabla} \times {\bf \hat{u}} = 0
\end{array} }\]
Applying the relation \({\bf \nabla} \times {\bf a} = \left( \frac{\partial a_z}{\partial y} - \frac{\partial a_y}{\partial z} \right)\hat{x} + \left( \frac{\partial a_x}{\partial z} - \frac{\partial a_z}{\partial x} \right) \hat{y} + \left( \frac{\partial a_y}{\partial x} - \frac{\partial a_x}{\partial y} \right) \hat{z}\)
to the \(x\)-component of (14) gives
(15)\[\begin{split} \renewcommand{\arraystretch}{2.0} \begin{array}{rcl}
\frac{\partial }{\partial y} \frac{ z }{ \left( x^2 + y^2 + z^2 \right)^{\frac{1}{2}} }
- \frac{\partial }{\partial z} \frac{ y }{ \left( x^2 + y^2 + z^2 \right)^{\frac{1}{2}} }
& = & \frac{- y z }{ \left( x^2 + y^2 + z^2 \right)^{\frac{3}{2}} }
- \frac{- y z }{ \left( x^2 + y^2 + z^2 \right)^{\frac{3}{2}} } = 0.
\end{array}\end{split}\]
The same pattern holds for the \(y\)- and \(z\)-components.
Laplacian of scalar field times unit vector
(16)\[\begin{split} \renewcommand{\arraystretch}{2.0} \begin{array}{rcl}
\nabla^2 c({\bf x}) \; {\bf \hat{u}}
& = & \left( \frac{\partial^2 c }{\partial x^2 } + \frac{\partial^2 c }{\partial y^2 } + \frac{\partial^2 c }{\partial z^2 } \right) \, {\bf \hat{u}} \\
& = & \frac{\partial }{\partial x } \left( \frac{\partial c }{\partial x } {\bf \hat{u}} \right)
+ \frac{\partial }{\partial y } \left( \frac{\partial c }{\partial y } {\bf \hat{u}} \right)
+ \frac{\partial }{\partial z } \left( \frac{\partial c }{\partial z } {\bf \hat{u}} \right) \\
& & - \left( \frac{\partial c }{\partial x } \frac{\partial {\bf \hat{u}}}{\partial x }
+ \frac{\partial c }{\partial y } \frac{\partial {\bf \hat{u}}}{\partial y }
+ \frac{\partial c }{\partial z } \frac{\partial {\bf \hat{u}}}{\partial z } \right) \\
& = & \nabla \cdot \left( \nabla c({\bf x}) \;{\bf \hat{u}} \right)
- \left( \frac{\partial c }{\partial x } \frac{\partial {\bf \hat{u}}}{\partial x }
+ \frac{\partial c }{\partial y } \frac{\partial {\bf \hat{u}}}{\partial y }
+ \frac{\partial c }{\partial z } \frac{\partial {\bf \hat{u}}}{\partial z } \right)
\end{array}\end{split}\]
Using (13)
(17)\[\begin{split} \renewcommand{\arraystretch}{2.0} \begin{array}{rcl}
-\left( \frac{\partial c }{\partial x } \frac{\partial {\bf \hat{u}}}{\partial x }
+ \frac{\partial c }{\partial y } \frac{\partial {\bf \hat{u}}}{\partial y }
+ \frac{\partial c }{\partial z } \frac{\partial {\bf \hat{u}}}{\partial z } \right)
& = & - \frac{1}{\rho} \left[ \frac{\partial c }{\partial x } \left( - u_x {\bf \hat{u}} + \hat{x} \right)
+ \frac{\partial c }{\partial y } \left( - u_y {\bf \hat{u}} + \hat{y} \right)
+ \frac{\partial c }{\partial z } \left( - u_z {\bf \hat{u}} + \hat{z} \right) \right] \\
& = & \frac{1}{\rho} \left( \nabla c \cdot {\bf \hat{u}} \right) {\bf \hat{u}}
- \frac{1}{\rho} \left( \nabla c \right)
\end{array}\end{split}\]
So
(18)\[\begin{split} \boxed{ \renewcommand{\arraystretch}{2.0} \begin{array}{rcl}
\nabla^2 c({\bf x}) \; {\bf \hat{u}}
& = & \nabla \cdot \left( \nabla c({\bf x}) \; {\bf \hat{u}} \right)
+ \frac{1}{\rho} \left( \nabla c \cdot {\bf \hat{u}} \right) {\bf \hat{u}}
- \frac{1}{\rho} \left( \nabla c \right)
\end{array} }\end{split}\]
Integral identities
Laplacian of scalar field integrated over sphere surface
Consider scalar field \(c({\bf x})\) defined on the surface of a sphere of radius \(\rho\) centered at the origin, point \((x,y,z)\) on the surface of the sphere, and \({\bf \hat{u}} =u_x\hat{x} + u_y\hat{y} +u_z \hat{z} = \frac{ x \hat{x} + y \hat{y} + z \hat{z} }{ \left( x^2 + y^2 + z^2 \right)^{\frac{1}{2}} }\) a the unit vector from the origin in the direction of the point \((x,y,z)\) on the sphere.
Show that the integral the Laplacian of scalar field over the surface of a sphere is zero.
(19)\[ \boxed{ \renewcommand{\arraystretch}{2.0} \begin{array}{rcl}
\int_{S} dS\, \nabla^2 c({\bf x}) = 0
\end{array} }\]
Applying (12) to (19)
gives the following components.
Since \(\frac{\partial c(\theta,\phi)}{\partial r}\).
(20)\[\begin{split} \renewcommand{\arraystretch}{1.0} \begin{array}{rcl}
\rho^2 \int_0^{2\pi} d\phi\, \int_0^{\pi}d\theta\, \sin\theta
\frac{1}{\rho^2} \frac{\partial }{\partial r}\left( r^2 \frac{\partial c(\theta,\phi)}{\partial r} \right) & = & 0,
\end{array}\end{split}\]
(21)\[\begin{split} \renewcommand{\arraystretch}{2.0} \begin{array}{rcl}
\rho^2 \int_0^{2\pi} d\phi\, \int_0^{\pi}d\theta\, \sin\theta
\frac{1}{\rho^2\sin\theta} \frac{\partial }
{\partial \theta} \left( \sin\theta \frac{\partial c(\theta,\phi) }{\partial \theta} \right)
& = & \int_0^{2\pi} d\phi\, \int_0^{\pi}d\theta\,
\frac{\partial }{\partial \theta} \left( \sin\theta \frac{\partial c(\theta,\phi)}{\partial \theta} \right) \\
& = & \int_0^{2\pi} d\phi\,
\left. \left( \sin\theta \frac{\partial c(\theta,\phi) }{\partial \theta} \right) \right|_0^{\pi} \\
& = & \int_0^{2\pi} d\phi\, \frac{\partial c(\pi,\phi) }{\partial \theta} \\
& = & \frac{\partial c(\pi,2\pi) }{\partial \theta} - \frac{\partial c(\pi,0) }{\partial \theta} = 0,
\end{array}\end{split}\]
and
(22)\[\begin{split} \renewcommand{\arraystretch}{1.0} \begin{array}{rcl}
\rho^2 \int_0^{\pi}d\theta\, \sin\theta \int_0^{2\pi} d\phi\,
\frac{1}{\rho^2\sin\theta} \frac{\partial^2 c(\theta,\phi) }{\partial \theta^2}
& = & \int_0^{\pi}d\theta\, \int_0^{2\pi} d\phi\, \frac{\partial^2 c(\theta,\phi) }{\partial \theta^2}
= \int_0^{\pi}d\theta\, \left. \frac{\partial c(\theta,\phi) }{\partial \theta} \right|_0^{2\pi}
= 0.
\end{array}\end{split}\]
So, \(\int_{S} dS\, \nabla^2 c({\bf x}) =0\).
Integral identity 1
Consider a vector \({\bf a}\), a sphere \(S\) of radius \(\rho\) centered at the origin, and a unit vector \({\bf \hat{u}}\) from the origin.
Show that
(23)\[\begin{split} \boxed{ \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\int_{S} dS \;\; \left( {\bf a} \cdot {\bf\hat{u}} \right) \hat{\bf u}
& = & \frac{4 \pi}{3} \rho^2 {\bf a}
\end{array} }\end{split}\]
In terms of spherical coordinates
(24)\[\begin{split} \renewcommand{\arraystretch}{1} \begin{array}{rcl}
{\bf \hat{u}} & = &
\sin(\phi) \cos(\theta) {\bf\hat{x}} - \sin(\phi)\sin(\theta) {\bf\hat{y}} - \cos(\phi) {\bf\hat{z}} \\
& = & \left[ \sin(\phi) \cos(\theta),\sin(\phi)\sin(\theta),\cos(\phi) \right]^T
\end{array}\end{split}\]
where we are using the convention that \(\theta\) is the azimuthal angle - the angle in the \(xy\)-plane (\(0 \le \theta \le 2\pi\)) and \(\phi\) is the polar (zenith) angle - the angle with respect to the \(z\)-axis (\(0 \le \phi \le \pi\)).
Then
(25)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\int_{S} dS \;\; \left( {\bf a} \cdot {\bf\hat{u}} \right) {\bf\hat{u}}
& = & \rho^2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \; {\bf a} \cdot {\bf\hat{u}} \;\;
\left[\sin(\phi) \cos(\theta),\sin(\phi)\sin(\theta),\cos(\phi) \right] \\
& \hspace{-50mm} = &\hspace{-25mm} \rho^2 \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\left( a_x \sin(\phi) \cos(\theta) + a_y \sin(\phi)\sin(\theta) + a_z \cos(\phi) \right)
\sin(\phi) \cos(\theta) \hat{\bf x} \\
& \hspace{-25mm} &\hspace{-25mm} + \rho^2 \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\left( a_x\sin(\phi) \cos(\theta) + a_y \sin(\phi)\sin(\theta) + a_z\cos(\phi) \right)
\sin(\phi)\sin(\theta) \hat{\bf y} \\
& \hspace{-25mm} &\hspace{-25mm} + \rho^2 \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\left( a_x\sin(\phi) \cos(\theta) + a_y \sin(\phi)\sin(\theta) + a_z\cos(\phi) \right)
\cos(\phi) \hat{\bf z} \\
& = & \rho^2 a_x\hat{\bf x} \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\sin^2(\phi) \cos^2(\theta) \\
& & + \rho^2 a_y \hat{\bf y} \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\sin^2(\phi)\sin^2(\theta) \\
& & + 2\pi \rho^2 a_z\hat{\bf z} \int_{0}^{\pi} d\phi \sin(\phi) \cos^2(\phi) \\
& = & \rho^2 a_x\hat{\bf x}
\left. \left( \frac{\theta}{2} + \frac{1}{4} \sin(2\theta) + \mbox{const} \right) \right|_{0}^{2\pi}
\int_{0}^{\pi} d\phi \sin^3(\phi) \\
& & + \rho^2 a_y
\left. \left( \frac{\theta}{2} - \frac{1}{4} \sin(2\theta) + \mbox{const} \right) \right|_{0}^{2\pi}
\hat{\bf y} \int_{0}^{\pi} d\phi \sin^3(\phi) \\
& & - 2\pi \rho^2 a_z\hat{\bf z} \left. \left( \frac{1}{3} \cos^3(\phi) \right) \right|_{0}^{\pi} \\
& = & \rho^2 a_x \frac{2\pi}{2}
\left. \left( \frac{\cos(3\phi)}{12} - \frac{3\cos(\phi)}{4} + \mbox{const} \right) \right|_{0}^{\pi} \hat{\bf x} \\
& & + \rho^2 a_y \frac{2\pi}{2}
\left. \left( \frac{\cos(3\phi)}{12} - \frac{3\cos(\phi)}{4} + \mbox{const} \right) \right|_{0}^{\pi} \hat{\bf y}
+ \frac{4}{3} \pi \rho^2 a_z\hat{\bf z} \\
& = & \rho^2 a_x \frac{2\pi}{2}
\left( \frac{-2}{12} - \frac{-6}{4} \right) \hat{\bf x}
+ \rho^2 a_y \frac{2\pi}{2}
\left( \frac{-2}{12} - \frac{-6}{4} \right) \hat{\bf y}
+ \frac{4 \pi}{3} \rho^2 a_z\hat{\bf z} \\
& = & \frac{4 \pi}{3} \rho^2 \left( a_x \hat{\bf x}
+ a_y \hat{\bf y}
+ a_z\hat{\bf z} \right) \\
& = & \frac{4 \pi}{3} \rho^2 {\bf a}
\end{array}\end{split}\]
Integral identity 2
For \(V\) a ball of radius \(\rho\) centered at the origin, vector \({\bf a}\), \({\bf x}\) a vector from the origin, and \({\bf \hat{n}} = \rho {\bf x}\) a unit vector from the origin.
Show that
(26)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\int_{V} dV \;\; {\bf a} \cdot {\bf x} \;\; \hat{\bf n}
& = &
\frac{4\pi}{15} \rho^5 {\bf a}
\end{array}\end{split}\]
In terms of spherical coordinates
(27)\[\begin{split} \renewcommand{\arraystretch}{1} \begin{array}{rcl}
{\bf \hat{n}} & = &
\sin(\phi) \cos(\theta) {\bf\hat{x}} - \sin(\phi)\sin(\theta) {\bf\hat{y}} - \cos(\phi) {\bf\hat{z}} \\
& = & \left[ \sin(\phi) \cos(\theta),\sin(\phi)\sin(\theta),\cos(\phi) \right]^T
\end{array}\end{split}\]
where we are using the convention that \(\theta\) is the azimuthal angle - the angle in the \(xy\)-plane (\(0 \le \theta \le 2\pi\)) and \(\phi\) is the polar (zenith) angle - the angle with respect to the \(z\)-axis (\(0 \le \phi \le \pi\)).
(28)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\int_{V} dV \;\; {\bf a} \cdot {\bf x} \;\; {\bf x}
& = & \int_{0}^{\rho} r^2 dr \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \;
{\bf a} \cdot r {\bf \hat{n} } \;\;
r \left[ −\sin(\phi) \cos(\theta),−\sin(\phi)\sin(\theta),−\cos(\phi) \right] \\
& \hspace{-50mm} = &\hspace{-25mm}
\int_{0}^{\rho} r^4 dr \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\left( a_x \sin(\phi) \cos(\theta) + a_y \sin(\phi)\sin(\theta) + a_z \cos(\phi) \right)
\sin(\phi) \cos(\theta) \hat{\bf x} \\
& \hspace{-25mm} &\hspace{-25mm} +
\int_{0}^{\rho} r^3 dr \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\left( a_x \sin(\phi) \cos(\theta) + a_y \sin(\phi)\sin(\theta) +a_z \cos(\phi) \right)
\sin(\phi)\sin(\theta) \hat{\bf y} \\
& \hspace{-25mm} &\hspace{-25mm}
+ \int_{0}^{\rho} r^3 dr \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\left( a_x \sin(\phi) \cos(\theta) + a_y \sin(\phi)\sin(\theta) +a_z \cos(\phi) \right)
\cos(\phi) \hat{\bf z} \\
& = & \int_{0}^{\rho} r^4 dr \, a_x \hat{\bf x} \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\sin^2(\phi) \cos^2(\theta) \\
& & +\int_{0}^{\rho} r^4 dr \, a_y \hat{\bf y} \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\sin^2(\phi)\sin^2(\theta) \\
& & + 2\pi \int_{0}^{\rho} r^4 dr \,a_z \hat{\bf z} \int_{0}^{\pi} d\phi \sin(\phi) \cos^2(\phi) \\
& = & \int_{0}^{\rho} r^4 dr \, a_x \hat{\bf x}
\left. \left( \frac{\theta}{2} + \frac{1}{4} \sin(2\theta) + \mbox{const} \right) \right|_{0}^{2\pi}
\int_{0}^{\pi} d\phi \sin^3(\phi) \\
& & + \int_{0}^{\rho} r^4 dr \, a_y
\left. \left( \frac{\theta}{2} - \frac{1}{4} \sin(2\theta) + \mbox{const} \right) \right|_{0}^{2\pi}
\hat{\bf y} \int_{0}^{\pi} d\phi \sin^3(\phi) \\
& & - 2\pi \int_{0}^{\rho} r^4 dr \,a_z \hat{\bf z} \left. \left( \frac{1}{3} \cos^3(\phi) \right) \right|_{0}^{\pi} \\
& = & \int_{0}^{\rho} r^4 dr \, a_x \pi
\left. \left( \frac{\cos(3\phi)}{12} - \frac{3\cos(\phi)}{4} + \mbox{const} \right) \right|_{0}^{\pi} \hat{\bf x} \\
& & + \int_{0}^{\rho} r^4 dr \, a_y \pi
\left. \left( \frac{\cos(3\phi)}{12} - \frac{3\cos(\phi)}{4} + \mbox{const} \right) \right|_{0}^{\pi} \hat{\bf y}
+ \frac{4}{3} \pi \int_{0}^{\rho} r^4 dr \, a_z \hat{\bf z} \\
& = & \frac{4 \pi}{3}\int_{0}^{\rho} r^4 dr \, \left( a_x \hat{\bf x}
+ a_y \hat{\bf y}
+a_z \hat{\bf z} \right) \\
& = & \frac{4\pi}{15} \rho^5 {\bf a}
\end{array}\end{split}\]
Integral identity 3
For \(V\) a ball of radius \(\rho\) centered at the origin, vector \({\bf a}\), \(a({\bf y})\) a scalar field, \({\bf y}\) a vector from the origin, \({\bf y({\bf x})} = {\bf x}\) a vector from the origin to the surface of the sphere, and the unit vector \({\bf \hat{x}}=\frac{1}{\rho} {\bf x}\).
Show that
(29)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\int_{V} dV \;\; \sum_i \partial_i \left(\partial_i a({\bf y})\; {\bf y} \right)
& = & \int_{S} dS \;\; \left[ {\bf y({\bf x})} {\bf \hat{x}} \cdot {\bf \nabla} a({\bf y({\bf x})}) \right]
\end{array}\end{split}\]
In terms of spherical coordinates
(30)\[\begin{split} \renewcommand{\arraystretch}{1} \begin{array}{rcl}
{\bf \hat{x}} & = &
\sin(\phi) \cos(\theta) {\bf\hat{x}} - \sin(\phi)\sin(\theta) {\bf\hat{y}} - \cos(\phi) {\bf\hat{z}} \\
& = & \left[ \sin(\phi) \cos(\theta),\sin(\phi)\sin(\theta),\cos(\phi) \right]^T
\end{array}\end{split}\]
where we are using the convention that \(\theta\) is the azimuthal angle - the angle in the \(xy\)-plane (\(0 \le \theta \le 2\pi\)) and \(\phi\) is the polar (zenith) angle - the angle with respect to the \(z\)-axis (\(0 \le \phi \le \pi\)).
(31)\[ \renewcommand{\arraystretch}{2} \begin{array}{rcl}
\nabla f = \frac{\partial f}{\partial r} \hat{r}
+ \frac{1}{r} \frac{\partial f}{\partial \theta} \hat{\theta}
+ \frac{1}{r\sin\theta} \frac{\partial f}{\partial \phi} \hat{\phi}
\end{array}\]
(32)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
I_1 & = & \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \int_{0}^{\rho} dr \; r^2
\frac{\partial }{\partial r} \left( \frac{\partial }{\partial r} a \; {\bf y} \right) \\ %\hat{r} \\
& = & \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \int_{0}^{\rho} dr \;
\frac{\partial }{\partial r} \left( r^2 \frac{\partial }{\partial r} a \; {\bf y} \right)
- 2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \int_{0}^{\rho} dr \;
r \frac{\partial }{\partial r} \left( \frac{\partial }{\partial r} a \; {\bf y} \right) \\
& = & \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \int_{0}^{\rho} dr \;
\frac{\partial }{\partial r} \left( r^2 \frac{\partial }{\partial r} a \; {\bf y} \right)
- 2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \int_{0}^{\rho} dr \;
\frac{\partial }{\partial r} \left( r \frac{\partial }{\partial r} a \; {\bf y} \right) \\
& & + 2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \int_{0}^{\rho} dr \;
\frac{\partial }{\partial r} \left( \frac{\partial }{\partial r} a \; {\bf y} \right) \\
& = & \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi)
\left. \left( r^2 \frac{\partial }{\partial r} a \; {\bf y} \right) \right|_0^\rho
- 2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi)
\left. \left( r \frac{\partial }{\partial r} a \; {\bf y} \right) \right|_0^\rho \\
& & + 2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \;
\left. \left( \frac{\partial }{\partial r} a \; {\bf y} \right) \right|_0^\rho \\
& = & \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi)
\left( \rho^2 \frac{\partial }{\partial r} a({\bf y({\bf x})}) \; {\bf y({\bf x})} \right)
- 2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi)
\left( \rho \frac{\partial }{\partial r} a({\bf y({\bf x})}) \; {\bf y({\bf x})} \right) \\
& & + 2 \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi) \;
\left( \frac{\partial }{\partial r} a({\bf y({\bf x})} ) \; {\bf y({\bf x})} \right) \\
& = & \int_{0}^{\pi} d\phi \int_{0}^{2\pi} d\theta \sin(\phi)
\left( \rho^2 - 2\rho + 2 \right) \frac{\partial }{\partial r} a({\bf y({\bf x})}) \; {\bf y({\bf x})}
\end{array}\end{split}\]
(33)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
I_2 & = & \int_{0}^{\rho} dr \; r^2 \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\frac{1}{r} \frac{\partial }{\partial \theta}
\left( \frac{1}{r} \frac{\partial a({\bf y})}{\partial \theta} \; {\bf y} \right) \\
& = & \int_{0}^{\rho} dr \; \int_{0}^{\pi} d\phi \sin(\phi) \int_{0}^{2\pi} d\theta
\frac{\partial }{\partial \theta}
\left( \frac{\partial a({\bf y})}{\partial \theta} \; {\bf y} \right) \\
& = & \int_{0}^{\rho} dr \; \int_{0}^{\pi} d\phi \sin(\phi)
\left. \left( \frac{\partial a({\bf y})}{\partial \theta} \; {\bf y} \right) \right|_{0}^{2\pi} \\
& = & \int_{S} d\!S \;
\left. \left( \frac{\partial a({\bf y})}{\partial \theta} \; {\bf y} \right) \right|_{0}^{2\pi}
\end{array}\end{split}\]
(34)\[\begin{split} \renewcommand{\arraystretch}{2} \begin{array}{rcl}
I_3 & = & \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \int_{0}^{\pi} d\phi \sin(\phi)
\frac{1}{\sin\theta} \frac{\partial }{\partial \phi}
\left(\frac{1}{r\sin\theta} \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \\
& = & \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{\sin^2\theta} \int_{0}^{\pi} d\phi \sin(\phi)
\frac{\partial }{\partial \phi}
\left( \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \\
& = & \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{\sin^2\theta} \int_{0}^{\pi} d\phi
\frac{\partial }{\partial \phi}
\left( \sin(\phi) \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \\
& & - \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{\sin^2\theta} \int_{0}^{\pi} d\phi
\cos(\phi) \frac{\partial }{\partial \phi}
\left( \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \\
& = & \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{r\sin^2\theta}
\left. \left( \sin(\phi) \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \right|_{0}^{\pi} \\
& & - \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{\sin^2\theta} \int_{0}^{\pi} d\phi
\frac{\partial }{\partial \phi}
\left( \cos(\phi) \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \\
& & - \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{\sin^2\theta} \int_{0}^{\pi} d\phi
\sin(\phi) \frac{\partial }{\partial \phi}
\left( \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \\
& = & - \frac{1}{2} \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{\sin^2\theta}
\left. \left( \cos(\phi) \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \right|_{0}^{\pi} \\
& = & \int_{0}^{\rho} dr \; r \int_{0}^{2\pi} d\theta \frac{1}{\sin^2\theta}
\left. \left( \frac{\partial }{\partial \phi} a({\bf y})\; {\bf y} \right) \right|_{0}^{\pi} \\
\end{array}\end{split}\]