I am new to multivariable calculus and I am trying to learn how surface area double integrals work. I am stuck on how to parameterize this function:
$\iint_S x dS$ where $S$ is the part of the cylinder $x^2 + 2y^2 = 2$ bounded between the planes $ z = -4$ and $z = y + 4x + 12$.
I first tried to convert to polar integrals and settled with this integral:
$\int_0^{\frac{\pi}{2}} \int_{-4} ^{z = y + 4x + 12} r \cos\theta \, r\, \mathrm dr \, \mathrm d\theta$ but I couldn't solve that integral.
So next I tried isolating x and y and making it in terms of $\mathrm dz\, \mathrm dx$, and I got
$\int_{-4} ^{y\sqrt{2-2y^2+12}} \int_{-\sqrt{2-2y^2}}^{\sqrt{-2y^2}} x \, \mathrm dx \, \mathrm dz$
Which I also couldn't solve.
Can anyone help point out what I am doing wrong here? This question is really stumping me!
The condition $x^2+2y^2=2$ is a very common type of condition, where a good ideal is to use modified cylindrical coordinates, by which I mean $$x=\sqrt{2}\cos{\theta},\quad y=\sin{\theta},\quad \theta \in \left[0,2\pi \right[.$$
Then, in lack of a better idea, to restrict $z$ to be in between $-4$ and $y+4x+12$, we can parametrize it as $$z= -4 + \left(y+4x+16\right)t = -4 + \left(\sin{\theta}+4\sqrt{2}\cos{\theta}+16\right)t, \quad t \in \left[0,1\right].$$
This gives us the parametrization $$\mathbf{r}(\theta,t)=(\sqrt{2}\cos{\theta},\sin{\theta},-4 + \left(\sin{\theta}+4\sqrt{2}\cos{\theta}+16\right)t),$$
which has the following partial derivatives $$\partial_\theta \mathbf{r}(\theta,t)=(-\sqrt{2}\sin{\theta},\cos{\theta},\left(\cos{\theta}-4\sqrt{2}\cos{\theta}\right)t),$$
$$\partial_t\mathbf{r}(\theta,t)=(0,0,\sin{\theta}+4\sqrt{2}\cos{\theta}+16).$$
Because of the form of these, it is simple enough to calculate their product, and as such, $$\left|\left| \partial_\theta \mathbf{r} \times \partial_t \mathbf{r}\right|\right| = \left(\sin{\theta}+4\sqrt{2}\cos{\theta}+16\right) \left|\left| (\cos{\theta},\sqrt{2}\sin{\theta},0)\right|\right| = \left(\sin{\theta}+4\sqrt{2}\cos{\theta}+16\right) \sqrt{1+\sin^2{\theta}} $$
This results in the integral being $$\int\int x dS = \int_0^1 \int_0^{2\pi} \sqrt{2}\cos{\theta} \left(\sin{\theta}+4\sqrt{2}\cos{\theta}+16\right) \sqrt{1+\sin^2{\theta}}\, d\theta \,dt,$$
which actually immediately reduces to a single-variable integral as nothing depends on $t$,
$$\int\int x dS = \sqrt{2} \int_0^{2\pi} \cos{\theta} \left(\sin{\theta}+4\sqrt{2}\cos{\theta}+16\right) \sqrt{1+\sin^2{\theta}}\, d\theta.$$
This is not an immediate integral, but we can make some some quick simplifications, namely by splitting it into three integrals. The first one, $$\int_0^{2\pi} \cos{\theta} \sin{\theta} \sqrt{1+\sin^2{\theta}}\, d\theta = 0,$$ is equal to zero, as we see with the change of variable $\theta \to 2\pi - \theta$ that it is equal to its symmetric. The third one,proportional to $$\int_0^{2\pi} \cos{\theta} \sqrt{1+\sin^2{\theta}}\, d\theta = 0,$$ is also zero, which we can quickly obtain splitting it into two integrals in $\left[0,\pi\right[$ and $\left[\pi,2\pi\right[$, and then performing the substitution $\theta\to \theta-\pi$ on the second one. Finally, only the middle term survives and we get $$\int\int x dS = 8\int_0^{2\pi} \cos^2{\theta} \sqrt{1+\sin^2{\theta}}\, d\theta.$$
With even more symmetries, you can reduce it to $$\int\int x dS = 32\int_0^{\frac{\pi}{2}} \cos^2{\theta} \sqrt{1+\sin^2{\theta}}\, d\theta.$$
The integrand function does not have a simple anti-derivative. This integral can be related to the Elliptic Integral, but it probably is not worth it to reduce it even further.