Deriving the surface area of a sphere using integration with spherical coordinates Say we have a sphere $x^2 + y^2 + z^2 = R^2$. Finding the volume of a sphere would be straight forward in spherical coordinates, since you would simply be integrating the domain, that is,
$$ V = \int_0^{\pi}\int_0^{2\pi}\int^R_{0} r^2\sin(\phi) \, dr d\theta d\phi \; \;\small(1)$$
Intuitively, since the radius remains a constant for just the surface area (i.e., you are not adding up the infinitesimally thin "layers" of the sphere, just the most outer one), the radius differential and integral is unnecessary, thus making the integral
$$ A = R^2\int_0^{\pi}\int_0^{2\pi} \sin(\phi) \, d\theta d\phi \; \;\small(2)$$
Geometrically, this result also makes sense. Since the differential representing the height of each "block" of the surface area would be $Rd\phi$, and the differential for the width would be $R\sin(\phi)d\theta$. Multiplying these two, we get $R^2\sin(\phi) \, d\theta d\phi$.
Is this intuition correct? Moreover, how would I go about deriving the second integral using the Jacobian?
 A: Consider a coordinate system $uvw$ with position vector $\vec{r}(u,v,w)$ on $\Bbb{R}^3$. Then the Jacobian will always defined by the scalar triple product
$$J(u,v,w) = \left|\frac{\partial\vec{r}}{\partial u}\cdot\left(\frac{\partial\vec{r}}{\partial v}\times\frac{\partial\vec{r}}{\partial w}\right)\right|$$
which is the formula for the volume of a parallel piped spanned by these vectors. Given the absolute values, the vectors of the partial derivatives can be put in any order in the above product.
In general, you would have to go through the entire surface integral process to calculate the Jacobian for an integral over an arbitrary surface. But consider the special case of $$\begin{cases}\text{(a) an orthogonal coordinate system, and}\\ \text{(b) integrating over a surface which is a constant in one of the variables, e.g. $u=k$}\\\end{cases}$$
Then, we have an easy formula for the surface Jacobian - it is given by:
$$dS = \left[\frac{J(u,v,w)}{\left|\frac{\partial\vec{r}}{\partial u}\right|}\right]_{u=k}\:dvdw$$
Heuristically, we can see that this reduces to the surface integral Jacobian formula because the $u$ partial derivatives "cancel out" (this step is made rigorous and justified by the condition that $uvw$ be an orthogonal coordinate system). Let's test this out on a few examples:
Spherical Coordinates $(\vec{r} = (r\sin\theta\cos\varphi,r\sin\theta\sin\varphi,r\cos\theta))$
Sphere $(r = R)$
$$\left[\frac{r^2\sin\theta}{\left|(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)\right|}\right]_{r=R} = \frac{R^2\sin\theta}{1} = R^2\sin\theta$$
Cone $(\theta=\alpha)$
$$\left[\frac{r^2\sin\theta}{\left|(r\cos\theta\cos\varphi,r\cos\theta\sin\varphi,-r\sin\theta)\right|}\right]_{\theta=\alpha} = \frac{r^2\sin\alpha}{r} = r\sin\alpha$$
Cylindrical Coordinates $(\vec{r} = (s\cos\varphi,s\sin\varphi,z))$
Cylinder $(s=R)$
$$\left[\frac{s}{|(\cos\theta,\sin\theta,0)|}\right]_{s=R} = \frac{R}{1} = R$$
And as an added bonus, if you were paying attention and noticed a pattern, you would find that we have a formula for finding the unit normal to a surface of the form $u=k$, which is needed to vector surface surface integrals. It is given by
$$\hat{n}_u = \frac{\frac{\partial\vec{r}}{\partial u}}{\left|\frac{\partial\vec{r}}{\partial u}\right|}$$
Combine both of these ideas and you can quickly do most common multivariable integral questions that appear on exams.
