# Find the surface area of cylinder between the intersection curves and cone, eliptic paraboloid.

The parametric equation of the eliptic paraboloid: $${r}(u,v) = \left(u\cos(v); u\sin(v); \dfrac{1}{4}u^2-\dfrac{21}{4}\right)\qquad 0 \leq v \leq 2\pi,\,\,0 \leq u \leq +\infty$$ The parametric equation of the cone: $${r}(u,v) = \left(u\cos(v); u\sin(v); -u\right)\qquad 0 \leq v \leq 2\pi,\,\,0 \leq u \leq +\infty$$ The parametric equation of the cylinder: $${r}(u,v) = \left(\cos(v)+1; \sin(v); u\right)\qquad 0 \leq v \leq 2\pi,\,\,-\infty \leq u \leq +\infty$$

I found that,

• The parametric equation of the intersection curve of the cylinder and the cone: $$\left(\cos(t)+1,\sin(t),\dfrac{1}{2}\cos(t)-\dfrac{19}{4}\right),\qquad 0\leq t\leq 2\pi$$ • The parametric equation of the intersection curve of the cylinder and the eliptic paraboloid: $$\left(\cos(t) + 1,\sin(t),-\sqrt{2 + 2cos(t)}\right),\qquad 0\leq t\leq 2\pi$$

But how to define the area of the cylinder limited by two curves above?

Enter the image of the decription

In Cartesian coordinates, the surfaces in question are given by the equations

$$\begin{cases} (x-1)^2 + y^2 = 1 & \text{(cylinder)} \\ z = -\sqrt{x^2+y^2} & \text{(cone)} \\ \frac{x^2}4 + \frac{y^2}4 = z+\frac{21}4 & \text{(ell.para.)} \end{cases}$$

From the cylinder's equation you immediately get

$$(x-1)^2+y^2=1 \implies x^2+y^2=2x$$

When the cylinder and cone meet, you have

$$z = -\sqrt{x^2+y^2} = -\sqrt{2x}$$

so that parameterizing $$x=\cos(v)+1$$ and $$y=\sin(v)$$ would yield $$z=-\sqrt{2\cos(v)+2}$$.

When the cylinder and elliptic paraboloid meet, you have

$$\frac{x^2+y^2}4 = z+\frac{21}4 \implies z = \frac x2-\frac{21}4$$

so that with $$x=\cos(v)+1$$ and $$y=\sin(v)$$, you get $$z=\frac{\cos(v)}2-\frac{19}4$$.

Then with $$z=u$$, the area of the surface you want is

$$\int_0^{2\pi} \int_{\tfrac{\cos(v)}2-\tfrac{19}4}^{-\sqrt{2\cos(v)+2}} \|r_u\times r_v\| \, du \, du$$

where $$r(u,v)$$ is the parameterization of the cylinder.