integrals for area in 3D space I'm in my first year of uni and we started using double and triple integrals in electrostatics to calculate the areas and volumes of simple shapes such as the cylinder. trying to learn the math behind it I learned about the jacobian and change of variables. now I understand why the triple integral for the volume of a cylinder is what it is since I can use the jacobian to get it from the one in cartesian.
$V(R)= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \chi(R)dxdydz$
to
$V(R)=\int_{0}^{2\pi}\int_{0}^{+\infty}\int_{-\infty}^{+\infty} \chi(R)\rho dzd\rho d\theta$
but the problem is when we calculated the area of the lateral surface of the cylinder in class using cylindrical coordinates.
$\int_{0}^{2\pi}\int_{0}^{h}R dzd\theta$
I didn't know what the integral would look like in cartesian, to begin with...I also saw some use the Dirac delta to calculate it in cartesian but I don't know how to manipulate those and I don't know how to transform it to cylindrical, my objective being to prove the use of the integrals the professor used.
QUESTION: can someone show me what the integral for the area of the lateral surface of a cylinder would look like in cartesian? and if it's complicated how to transform it into cylindrical.
 A: Integral in cartesian
One way to find the lateral surface area is to add up the areas of
thickened circles of small height like "$\mathrm{d}z$". To do
that, we first need the circumference of those circles.
The circumference of a circle $x^{2}+y^{2}=R^{2}$ (at height $z$)
doesn't change with height, so we can just work in the $xy$-plane.
Then there are two semicircles, one given by $y=\sqrt{R^{2}-x^{2}}$
and the other given by $y=-\sqrt{R^{2}-x^{2}}$. The semicircles each
have the same length, so we can just find twice the length of the
first one. The arc length formula can be found at Paul's Online Notes or
or openstax Calculus Volume 2. We have $\int_{-R}^{R}\sqrt{1+\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\sqrt{R^{2}-x^{2}}\right)^{2}}\,\mathrm{d}x$
for the length of one semicircle, so twice that is the entire circumference.
Then we can multiply these circumferences by a small height $\Delta z$,
add them up, and take a limit to turn it into the following integral:
\begin{align*}
 & \int_{0}^{h}\left(2\int_{-R}^{R}\sqrt{1+\left(\dfrac{\mathrm{d}}{\mathrm{d}x}\sqrt{R^{2}-x^{2}}\right)^{2}}\,\mathrm{d}x\right)\,\mathrm{d}z\\
= & 2\int_{0}^{h}\int_{-R}^{R}\sqrt{1+\left(-\dfrac{x}{\sqrt{R^{2}-x^{2}}}\right)^{2}}\,\mathrm{d}x\,\mathrm{d}z\\
= & 2\int_{0}^{h}\int_{-R}^{R}\sqrt{1+\dfrac{x^{2}}{R^{2}-x^{2}}}\,\mathrm{d}x\,\mathrm{d}z\\
= & 2\int_{0}^{h}\int_{-R}^{R}\sqrt{\dfrac{R^{2}}{R^{2}-x^{2}}}\,\mathrm{d}x\,\mathrm{d}z\\
= & 2\int_{0}^{h}\int_{-R}^{R}\dfrac{R}{\sqrt{R^{2}-x^{2}}}\,\mathrm{d}x\,\mathrm{d}z\\
= & 2\int_{0}^{h}\int_{-R}^{R}\dfrac{1}{\sqrt{1-\left(x/R\right)^{2}}}\,\mathrm{d}x\,\mathrm{d}z
\end{align*}
Note that the inner $x$ integral doesn't depend on $z$, so we could
also write
$$
2\left(\int_{-R}^{R}\dfrac{1}{\sqrt{1-\left(x/R\right)^{2}}}\,\mathrm{d}x\right)\left(\int_{0}^{h}1\,\mathrm{d}z\right)
$$
$$
\text{or }\int_{-R}^{R}\left(\dfrac{1}{\sqrt{1-\left(x/R\right)^{2}}}2\int_{0}^{h}1\,\mathrm{d}z\right)\,\mathrm{d}x\text{.}
$$
The latter suggests a way of thinking of this by cutting slices for
each value of $x$, where the $\dfrac{1}{\sqrt{1-\left(x/R\right)^{2}}}$
accounts for the curve of the semicircle.
Transforming back to cylindrical
To transform back to cylindrical, we can use a trigonometric substitution
(openstax, Paul's). We use $x=R\cos\theta$ and $\mathrm{d}x=-R\sin\theta\,\mathrm{d}\theta$
to convert the $x$ integral above:
\begin{align*}
 & \int_{-R}^{R}\dfrac{1}{\sqrt{1-\left(x/R\right)^{2}}}\,\mathrm{d}x\\
= & \int_{\theta\text{ when }x=-R}^{\theta\text{ when }x=R}\dfrac{1}{\sqrt{1-\left(R\cos\theta/R\right)^{2}}}\left(-R\sin\theta\right)\,\mathrm{d}\theta\\
= & \int_{\pi}^{0}\dfrac{1}{\sqrt{1-\cos^{2}\theta}}\left(-R\sin\theta\right)\,\mathrm{d}\theta\\
= & \int_{\pi}^{0}\dfrac{-R\sin\theta}{\sqrt{\sin^{2}\theta}}\,\mathrm{d}\theta\\
= & \int_{\pi}^{0}\dfrac{-R\sin\theta}{\left|\sin\theta\right|}\,\mathrm{d}\theta\\
= & \int_{\pi}^{0}-R\,\mathrm{d}\theta\text{ since }\sin\theta\ge0\text{ on }[0,\pi]\\
= & \int_{0}^{\pi}R\,\mathrm{d}\theta
\end{align*}
This means the entire area becomes
$$
2\left(\int_{0}^{\pi}R\,\mathrm{d}\theta\right)\left(\int_{0}^{h}1\,\mathrm{d}z\right)=\left(\int_{0}^{2\pi}R\,\mathrm{d}\theta\right)\left(\int_{0}^{h}1\,\mathrm{d}z\right)=\boxed{\int_{0}^{2\pi}\int_{0}^{h}R\,\mathrm{d}z\,\mathrm{d}\theta}
$$
More general areas
The lateral area of a cylinder is a very special kind of surface,
and there are more general techniques we can use to find the area
of it and other surfaces.
Areas of revolution
The cylinder is a surface of revolution: For the cylinder under discussion,
we revolve the line segment $x=R,y=0,z\in[0,h]$ around the $z$-axis.
Rotating things around, this would have the same area as rotating
the line segment $x=R$, $y\in[0,h]$ (in the plane $z=0$) around
the $y$-axis. The standard formula (Paul's, openstax) for the lateral area of a surface of revolution gives:
\begin{align*}
 & \int_{0}^{h}2\pi R\sqrt{1+\left(\dfrac{\mathrm{d}}{\mathrm{d}y}R\right)^{2}}\,\mathrm{d}y\\
= & 2\pi\int_{0}^{h}R\,\mathrm{d}y\\
= & \left(\int_{0}^{2\pi}1\,\mathrm{d}\theta\right)\int_{0}^{h}R\,\mathrm{d}y\\
= & \int_{0}^{2\pi}\int_{0}^{h}R\,\mathrm{d}y\,\mathrm{d}\theta\\
= & \int_{0}^{2\pi}\int_{0}^{h}R\,\mathrm{d}z\,\mathrm{d}\theta
\end{align*}
Areas of parametrized surfaces
If you do more calculus (you might not come across this in a
first study of electrostatics), you might want to find the surface
area of a surface that doesn't have any rotational symmetry, but still
has a nice parametrization. This is described at openstax Calculus Volume 3 and Paul's Online Calculus III Notes, for instance.
One thing to watch out for is that if you have something like a Moebius
strip which has only one side, you might not be calculating the area
you intended. See the Math StackExchange question "Area of Mobius strip" and its answers for some discussion.
