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

Question

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

Answer 1

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.