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I'm trying to draw the region of the surface area of the cylinder, $x^2+y^2 \le 2x$, limited by the cone $z=\sqrt{(x^2+y^2)}$ and the plane $z=0$ . I know that the cylinder's center is at $(1,0)$ $x^2+y^2\leq2x$ can be rewritten as $(x-1)^2+y^2\leq1$: a closed circle with center $(1,0)$ and radius $1$.

First parametrise by setting $x-1=\cos \theta$, $y=\sin \theta$

This gives $z=\sqrt{2x}=\sqrt{2(\cos \theta+1)}$.

We can now imagine the cylinder being unwrapped, so that the area is:

$$\int_0^{2\pi} \sqrt{2(\cos \theta+1)} d\theta$$

I have no idea how to visualize the part of the cylinder $x^2+y^2 \le 2x$ limited by the cone $z=\sqrt{(x^2+y^2)}$ and the plane $z=0$.

Trying to visualize, I'm guessing that the surface area given by the integral is the area

$$\int_0^{2\pi} \sqrt{2(\cos \theta+1)} d\theta$$

of a conic section?

I don’t think the cylinder is bounded above which is a problem.

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  • $\begingroup$ Is surface area a part of the surface $ z=\sqrt{2x}$ ? $\endgroup$ – user515045 Jan 2 '18 at 16:32
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The cylinder is tangent to the $y$ axes in $(0,0)$, which is the cone vertex. Of the cone you are considering the upper part $0 \le z$.
The portion of cylinder below that is a sort of "nail".

Working in cylindrical coordinates $(r, \phi, z)$ centered at $(1,0,0)$, the point $(r=1, -\pi < \phi \le \pi, 0)$ will be distant $\sqrt{(1+cos\phi)^2+sin^2\phi}$ from the origin, and the cone will be at the same distance above it.

Along the base circle $dl=d\phi$ and the elementary surface is $dA= \sqrt{(1+cos\phi)^2+sin^2\phi}\;d\phi$.

Cone_Cyl_1

Can you continue from here ?

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  • $\begingroup$ No, I still cant visualize whats going on $\endgroup$ – user515045 Jan 2 '18 at 17:21
  • $\begingroup$ I can’t see how the region is bounded above. Of course the region is bounded below by z=0. $\endgroup$ – user515045 Jan 2 '18 at 17:24
  • $\begingroup$ Can you draw a picture? $\endgroup$ – user515045 Jan 2 '18 at 17:25
  • $\begingroup$ Can you visualize a upward circular cone with vertex in the origin ? and $45^\circ$ aperture ? a sort of big icecream ? Can you visualize to lay on the ground a metal ring of radius $1$, and push it to touch the vertex ? Can you then visualize to raise from the circle a vertical curtain / fence till touching the cone above ? that starts at $0$ at the vertex, goes gradually increasing untill you reach the point on the ring diametrically opposite to the vertex, which is distant $2$ from the vertex, and so where the curtain reaches its max height = $2$? and symmetrically for the other half ? $\endgroup$ – G Cab Jan 2 '18 at 17:44
  • $\begingroup$ Are we pushing the metal ring of radius 1 up in the z-direction? Is the metal ring of radius one only touching the vertex only at one point? $\endgroup$ – user515045 Jan 2 '18 at 17:51

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