In spherical polars, $$x=r\cos(\phi)\sin(\theta)$$ $$y=r\sin(\phi)\sin(\theta)$$ $$z=r\cos(\theta)$$ I want to work out an integral over the surface of a sphere - ie $r$ constant. I'm able to derive through scale factors, ie $\delta(s)^2=h_1^2\delta(\theta)^2+h_2^2\delta(\phi)^2$ (note $\delta(r)=0$), that: $$h_1=r\sin(\theta),h_2=r$$ $$dA=h_1h_2=r^2\sin(\theta)$$

I'm just wondering is there an "easier" way to do this (eg. Jacobian determinant when I'm varying all 3 variables). I know you can supposedly visualize a change of area on the surface of the sphere, but I'm not particularly good at doing that sadly.


6 Answers 6


I've come across the picture you're looking for in physics textbooks before (say, in classical mechanics). A bit of googling and I found this one for you!area element on sphere

Alternatively, we can use the first fundamental form to determine the surface area element. Recall that this is the metric tensor, whose components are obtained by taking the inner product of two tangent vectors on your space, i.e. $g_{i j}= X_i \cdot X_j$ for tangent vectors $X_i, X_j$. We make the following identification for the components of the metric tensor, $$ (g_{i j}) = \left(\begin{array}{cc} E & F \\ F & G \end{array} \right), $$ so that $E = <X_u, X_u>, F=<X_u,X_v>,$ and $G=<X_v,X_v>.$

We can then make use of Lagrange's Identity, which tells us that the squared area of a parallelogram in space is equal to the sum of the squares of its projections onto the Cartesian plane: $$|X_u \times X_v|^2 = |X_u|^2 |X_v|^2 - (X_u \cdot X_v)^2.$$
Here's a picture in the case of the sphere:

Projecting surface element onto the Cartesian plane

This means that our area element is given by $$ dA = | X_u \times X_v | du dv = \sqrt{|X_u|^2 |X_v|^2 - (X_u \cdot X_v)^2} du dv = \sqrt{EG - F^2} du dv. $$

So let's finish your sphere example. We'll find our tangent vectors via the usual parametrization which you gave, namely, $X(\phi,\theta) = (r \cos(\phi)\sin(\theta),r \sin(\phi)\sin(\theta),r \cos(\theta)),$ so that our tangent vectors are simply $$ X_{\phi} = (-r\sin(\phi)\sin(\theta),r\cos(\phi)\sin(\theta),0), \\ X_{\theta} = (r\cos(\phi)\cos(\theta),r\sin(\phi)\cos(\theta),-r\sin(\theta)) $$ Computing the elements of the first fundamental form, we find that $$ E = r^2 \sin^2(\theta), \hspace{3mm} F=0, \hspace{3mm} G= r^2. $$ Thus, we have $$ dA = \sqrt{r^4 \sin^2(\theta)}d\theta d\phi = r^2\sin(\theta) d\theta d\phi $$

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    $\begingroup$ Yea I saw that too, I'm just wondering if there's some other way similar to using Jacobian (if someday I'm asked to find it in a self-invented set of coordinates where I can't picture it) $\endgroup$
    – Rcwt
    Commented Apr 14, 2012 at 15:10
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    $\begingroup$ @R.C. I've edited my response for you. Perhaps this is what you were looking for... ? $\endgroup$ Commented Apr 26, 2012 at 4:58

When you have a parametric representatuion of a surface $$S:\quad (u,v)\ \mapsto\ {\bf x}(u,v)$$ then an infinitesimal rectangle $[u, u+du]\times [v,v+dv]$ in the parameter plane is mapped onto an infinitesimal parallelogram $dP$ having a vertex at ${\bf x}(u,v)$ and being spanned by the two vectors ${\bf x}_u(u,v)\, du$ and ${\bf x}_v(u,v)\,dv$. The area of this parallelogram is $${\rm d}\omega:=|{\bf x}_u(u,v)\times{\bf x}_v(u,v)|\ {\rm d}(u,v)\ .$$ If you are given a "surface density“ ${\bf x}\mapsto \rho({\bf x})$ $\ ({\bf x}\in S)$ then the integral $I(S)$ of this density over $S$ is then given by $$I(S)=\int_B \rho\bigl({\bf x}(u,v)\bigr)\ {\rm d}\omega = \int_B \rho\bigl({\bf x}(u,v)\bigr)\ |{\bf x}_u(u,v)\times{\bf x}_v(u,v)|\ {\rm d}(u,v)\ ,$$ where $B$ is the parameter domain corresponding to the exact piece $S$ of surface.

Now this is the general setup. When your surface is a piece of a sphere of radius $r$ then the parametric representation you have given applies, and if you just want to compute the euclidean area of $S$ then $\rho({\bf x})\equiv1$.

Therefore in your situation it remains to compute the vector product ${\bf x}_\phi\times {\bf x}_\theta$ as a function of $\phi$ and $\theta$, resp., the absolute value of this product, and then you have to integrate over the desired parameter domain $B$.

You have explicitly asked for an explanation in terms of "Jacobians". The vector product $\times$ is the appropriate surrogate of that in the present circumstances, but in the simple case of a sphere it is pretty obvious that ${\rm d}\omega=r^2\sin\theta\,{\rm d}(\theta,\phi)$.


There is yet another way to look at it using the notion of the solid angle. Then the area element has a particularly simple form: $$dA=r^2d\Omega$$

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    $\begingroup$ Shouldn't be $r^2$? $\endgroup$ Commented Jul 31, 2015 at 13:33
  • $\begingroup$ Remember that the area asociated to the solid angle is given by $A=r^2 \Omega $ en.wikipedia.org/wiki/Solid_angle $\endgroup$ Commented Oct 11, 2018 at 22:21
  • $\begingroup$ fixed, thanks for pointing out $\endgroup$
    – Valentin
    Commented Oct 13, 2018 at 21:53

Because of the striking fact that

(a) The area of [a slice of the spherical surface between two parallel planes (within the poles)] is proportional to its width.

. . . here's a rarely (if ever) mentioned way to integrate over a spherical surface. We assume the radius = 1.

(b) Note that every point on the sphere is uniquely determined by its z-coordinate and its counterclockwise angle phi, $0 \leq\phi\leq 2\pi$, from the half-plane y = 0, x >= 0.

From (a) and (b) it follows that an element of area on the unit sphere centered at the origin in 3-space is just dphi dz.

Then the integral of a function f(phi,z) over the spherical surface is just $$\int_{-1 \leq z \leq 1, 0 \leq \phi \leq 2\pi} f(\phi,z) d\phi dz$$.


The answers above are all too formal, to my mind. The straightforward way to do this is just the Jacobian. The Jacobian is the determinant of the matrix of first partial derivatives. The first row is $\partial r/\partial x$, $\partial r/\partial y$, etc, the second the same but with $r$ replaced with $\theta$ and then the third row replaced with $\phi$. So to compute each partial you hold the other variables constant and just differentiate with respect to the variable in the denominator, e.g. $r=\sqrt{x^2+y^2+z^2}$

so $\partial r/\partial x = x/r $. You then just take the determinant of this 3-by-3 matrix, which can be done by cofactor expansion for instance.


There is an intuitive explanation for that.

Warm up

We are trying to integrate the area of a sphere with radius r in spherical coordinates. The angle $\theta$ runs from the North pole to South pole in radians. Angle $\theta$ equals zero at North pole and $\pi$ at South pole. The distance on the surface of our sphere between North to South poles is $r \, \pi$ (half the circumference of a circle).

Lines on a sphere that connect the North and the South poles I will call longitudes.

In order to calculate the area of a sphere we cover its surface with small RECTANGLES and sum up their total area.


In baby physics books one encounters this expression. In each infinitesimal rectangle the longitude component is its vertical side. The latitude component is its horizontal side.

$$\int_{0}^{ \pi }\int_{0}^{2 \pi } r^2 \sin {\theta} \, d\phi \,d\theta = \int_{0}^{ \pi }\int_{0}^{2 \pi } \overbrace{ \underbrace {r \, d\theta}_{\text{longitude component}} *\underbrace {r \, \color{blue}{\sin{\theta}} \,d \phi}_{\text{latitude component}}}^{\text{area of an infinitesimal rectangle}} $$

We see that the latitude component has the $\color{blue}{\sin{\theta}}$ adjustment to it. This is key. Why is that?

It is because rectangles that we integrate look like ordinary rectangles only at equator! Where $\color{blue}{\sin{\frac{\pi}{2}} = 1}$, i.e. where we do not need to adjust the latitude component. Near the North and South poles the rectangles are warped. Here is the picture.

enter image description here

The blue vertical line is longitude 0. The brown line on the right is the next longitude to the east. Blue triangles, one at each pole and two at the equator, have markings on them. These markings represent equal angles for $\theta \, \text{and} \, \phi$.

We need to shrink the width (latitude component) of integration rectangles that lay away from the equator. Because only at equator they are not distorted. Why we choose the sine function? Intuitively, because its value goes from zero to 1, and then back to zero. That is, where $\theta$ and radius $r$ map out the zero longitude (part of a circle of a plane).

What happens when we drop this sine adjustment for the latitude?

Case B: drop the sine adjustment for the latitude

In this case all integration rectangles will be regular undistorted rectangles. Their total length along a longitude will be $r \, \pi$ and total length along the equator latitude will be $r \, 2\pi$. Total area will be $$r \, \pi \times r \, 2\pi = 2 \pi^2 \, r^2$$

Like this $$\int_{0}^{ \pi }\int_{0}^{2 \pi } r \, d\theta * r \, d \phi = 2 \pi^2 r^2$$

Here is the picture.

enter image description here


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