Question

Hi.

I am currently doing calculus 2 and came across a problem I cannot get right.
There is a circular light beam with radius of 20 shining on a tank with radius 40.

I am required to calculate the area of the tank that is covered by light.

I have tried solving arc length for the cylindrical surface making an elliptical equation with small radius = beam radius and a big radius = the arc length of the tank as shown in the drawing.

This gives me a good estimate, but I am told that i need to set up the surface integral
https://i.gyazo.com/cd8dcdfda5f1a4c5ea75c3d7e58271ca.png

Answer 1

my estimate is 1315 while the actual answere is 1300.xx

Answer 2

@fixxer This guy, HolsterEmission, helping you now is a mathematical genius. I know he can help you. The problem will be trying to understanding his brilliant work!

Answer 3

Cant wait, i have been trying setting it up for both polar and Cartesian coordinates, I get it wrong every time :(.

Glad there is professionals around to support us who is not that brilliant :)

Answer 4

Let's say the tank is modeled by the cylindrical surface \(x^2+y^2=40^2\), so it's an upright tank with its base in the \(xy\) plane.

The beam of light can also be considered a cylinder, assuming light doesn't radiate outward from the source. It's pointing in a perpendicular direction, so let's just choose an axis. Say it travels along the \(y\) axis, so this cylinder is given by \(x^2+z^2=20^2\).

We can find a parameterization for the intersection of the smaller cylinder with the larger one. Solving for \(x\) in both equations, you get
\[\begin{cases}
x^2+y^2=40^2\\[1ex]x^2+z^2=20^2\end{cases}\implies\begin{cases}y=\pm\sqrt{40^2-x^2}\\[1ex]z=\pm\sqrt{20^2-x^2}\end{cases}\]omitting the negative root for \(y\) so that we only consider one side of the larger cylinder.

It took some playing around with, but I managed to find a parameterization for the surface bounded by the intersection, given by \(\vec{r}(x,t)=(x\cos t,\sqrt{40^2-x^2},\pm\sqrt{20^2-x^2})\), where \(|x|\le40\) and \(0\le t\le \dfrac{\pi}{2}\). I'm not sure why this works (not a lot of experience of parameterizing surfaces), but it does. I've attached a visual.

The orange cylinder is the tank, the green cylinder represents the unhindered beam of light, the red contour (it's not a circle, but it might seem that way) is the intersection of the two cylinders, and the blue region with the white mesh is the part of the surface of the tank where the light hits it.

Answer 5

So if we call this blue region \(S\), we should be able to find the area of \(S\) by computing
\[\iint_S \mathrm d\Sigma=\int_{-40}^{40} \int_0^{\pi/2} \left\|\frac{\partial \vec{r}}{\partial x}\times\frac{\partial \vec{r}}{\partial t}\right\|\,\mathrm dt\,\mathrm dx\]I have yet to check if this is doable, but if it's not it'll be because the surface parameterization isn't ideal.

Answer 6

A slightly simpler parameterization would actually be
\[\vec{r}(x,t)=(xt,\sqrt{40^2-x^2},\pm\sqrt{20^2-x^2})\]with \(0\le t\le1\).

Answer 7

Thank you allot for all your help, im trying to evaluate the integral now, it seems to make sense the way you put it.

Might finish tomorrow it is getting late here :p

Answer 8

Hope this works out, if not I'll need a serious Calc III review...

Answer 9

Hi, I ended up solving the surface integral this way, thank you allot for your help Holster :)
You are one educated man!