Finding the cost of flat object with mathematically defined shape

Question

I am given the following problem:

You wish to manufacture a zinc object with the shape of the surface of the cylinder $x^2 + y^2 = 4$ between the planes $z = 0$ and $x+y+z = 2$ (with $z \geq 0$). If the square meter of zinc cost $M$, what is the object's price?

My approach was to evaluate the volume of the object:

\begin{align*} V = \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{2-r\cos \theta - r \sin \theta} \ r dzdrd\theta = \int_{0}^{2\pi} \int_{0}^{2} (2r - r^2 \cos \theta - r^2 \sin \theta ) \ drd\theta = \cdots = 8 \pi \end{align*}

but I'm pretty sure that was useless, by the fact that the material cost is not related to the volume, but the surface area.

Can anyone point me into a direction?

Thank you.

Answer 1

It may be useful to consider the change of coordinates by rotating the $xy$-plane by $45^\circ$ as: $$u = x+y,\; v=x-y.$$ Then the top surface of the cylinder is described by the simple equation $z=2-u$ and you can calculate the elliptical surface area in terms of principle axes.

I hope, this is helpful.

Answer 2

You can change the order of integration of your volume so that it is $$ V = \int_0^2 \int_0^{2\pi} \int_0^{2-r\cos \theta - r \sin \theta} r \,dz\,d\theta\,dr. $$

Notice that this is like the shell method of integrating a volume of revolution, except that (because it is not a volume of revolution) the innermost integral has bounds that depend on $\theta.$ Despite this difference from the usual shell method, it is still the case that the outer integral is integrating over concentric "shells," each of which is some part of a cylinder. You can write the integral as $$ V = \int_0^2 A(r)\,dr $$ where $A(r)$ is the area of the shell that has radius $r.$

Notice that the part of a cylinder whose area you are looking for is also the outermost shell of this integral (the shell with radius $2$). That should be a big hint about what integral would represent the area of that part of a cylinder.

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But there is another wrinkle in this problem that one might miss. If $\theta = 0$ or if $\theta = \frac\pi2,$ then $2-r\cos \theta - r \sin \theta = 0.$ It turns out that $2-r\cos \theta - r \sin \theta < 0$ whenever $0 < \theta < \frac\pi2.$

So there are actually two pieces of the cylinder $x^2 + y^2 = 4$ between the planes $z = 0$ and $x + y + z = 2$: one above the plane $z = 0$ and one below that plane. If we just blindly integrated using the bound $2-r\cos \theta - r \sin \theta$ over the entire interval $\theta = 0$ to $\theta = 2\pi,$ we would count the piece below the plane as negative area.

I have never heard of negative zinc, so I suppose we are not allowed to do this. The reasonable alternatives are to negate the area below the plane $z = 0$ so that it is treated as positive area, or leave out that part of the cylinder's surface entirely. Since the problem statement says $z \geq 0,$ my interpretation would be that we are supposed to leave out the part of the cylinder below the plane $z = 0.$ In order to do that, the bounds of integration of $\theta$ have to be adjusted so that they do not include the interval $\left[0,\frac\pi2\right]$.