Find the area of an intersection of two curves, where one of them is missing a component

Question

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.

Answer 1

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.

Answer 2

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.