2

I have this problem asking to find the area of the intersection of $x^2+y^2=2x$ (cylinder centered at x=1) and $z=\sqrt{x^2+y^2}$ (positive branch of a cone)

I have two problems: since one of the surfaces doesn't have a z component, I'm not sure how to find the intersection. Otherwise I would solve for z and set both equations as equal, or replace with the value of z in one of them. But this time I'm lost.

And then my second problem is: if I don't get it wrong, this intersection gives me a curve (which will be a circle, I believe). So all I should do after I find the equation for such curve is a double integral to get its area, right?

Thanks.

EDIT: this doesn't seem to be a duplicate of Finding surface area of cone inside a cylinder since I'm trying to find the area of a curve and not a surface area, like in the other problem.

Floella
  • 473
  • Possible duplicate of Finding surface area of cone inside a cylinder – Intelligenti pauca Dec 14 '17 at 20:49
  • If $z$ doesn’t appear in the equation, that means that it’s unconstrained: any value of $z$ will satisfy the equation. Solve for some other variable instead. Observe, too, that $x^2+y^2$ appears in both equations. You might be able to do something with that. – amd Dec 14 '17 at 21:19
  • Not the exact same problem as the linked one, I think, as in that case the surface area of the cone is being found. In this case, it's the area of the curve intersection of two surfaces. I think I need the projection in the xy plane of this curve, and then get the area using polar coordinates maybe, but I'm not sure of how to find that projection on the xy plane... I did set $x^2+y^2=r^2$ and then $z=r$ but that's as far as I got. – Floella Dec 14 '17 at 22:06
  • A curve has zero area. There, you are done without any further calculation. (Or perhaps there is a mistake in the problem statement.) – David K Dec 14 '17 at 23:16
  • Hmm... I thought I wanted to find the area of the circle that is an intersection of both surfaces. Doesn't it have an area? – Floella Dec 14 '17 at 23:26

2 Answers2

0

Let me see if I got it right... In order to find the intersection of the surfaces I need to trace both in the xy plane, for which I need to set $z=0$. In the cylinder I have no z, so I'd just trace a circle of radius 1, centered at $x=1$. For the other one, I'd have $0=\sqrt{x^2+y^2}$ so that would be just the point $(0,0)$. So this means the projection in the xy plane is the circle of radius 1 centered at $x=1$. If I use polar coordinates and integrate $r$ from 0 to 1 and $\theta$ from 0 to $2\pi$ I'd have the area I'm looking for. Please correct me if I'm wrong.

Floella
  • 473
0

You have $z=f(x,y)=\sqrt{x^2+y^2}$ with $f_x=\frac{x}{\sqrt{x^2+y^2}}$ and $f_y=\frac{y}{\sqrt{x^2+y^2}}$. Then the surface area of $S$ is given by $$\displaystyle\int\int_R\sqrt{1+f_x^2+f_y^2}dA,$$ where $R$ is the region enclosed by the circle centered at $(1,0)$: $(x-1)^2+y^2=1$, obtained by completing a square of $x^2+y^2=2x$. After a computation, we have $$\displaystyle\int\int_R\sqrt{1+f_x^2+f_y^2}dA=\displaystyle\int\int_R\sqrt{2}dA.$$ Now to evaluate the integral you can use polar coordinates. Using polar coordinates you get $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$ and $0\leq r\ \leq 2\cos\theta$.

Thus evaluation of the integral $$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{2\cos\theta}\sqrt{2}rdrd\theta$$ will give you the result.

daulomb
  • 3,955