2
$\begingroup$

calculate the surface area of the part of a cylinder $ x^2 + (y-1)^2 = 1 $ that is inside the sphere $ x^2 + y^2 + z^2 = 4 $.

my trial :

The Domain of integration on the YZ plane is :

solving :

(*) $ x^2 + (y-1)^2 = 1 $

$ x^2 + (y)^2 + z^2 \leq 4 $

We get : $ 0 \leq z \leq \sqrt{4-2y}$ and $ -2 \leq y \leq 2 $

$||\nabla{Cylinder(x,y,z)}|| = ||(2x,2(y-1),0)|| = 2 ~$ see(*)

S = $2~\int_{-2}^{2}\int_{0}^{\sqrt{4-2y}}~~2~dydz$ ( by symmetry *2)

I also tried solving by parmetrization and i didn't get the same answer i really need HELP is this way alright ? or is there something wrong

$\endgroup$
2
$\begingroup$

Let $\gamma : [0, 2\pi] \to \mathbb{R}^2, \gamma(t) = (\cos t, 1+\sin t)$ be the parameterization of the circle $x^2+(y-1)^2 = 1$ in the $xy$-plane.

The height of the cylinder over a point $(x,y)$ on this circle is given by $z = \sqrt{4-x^2-y^2}$ so the area is

$$A = 2\int_\gamma z\,d\gamma = 2\int_\gamma \sqrt{4-x^2-y^2}\,d\gamma = 2\int_0^{2\pi}\sqrt{4-\cos^2t -(1+\sin t)^2}\,dt = 2\int_0^{2\pi}\sqrt{2(1-\sin t)}\,dt$$

We can solve this integral by noting that $1-\sin t = \left(\sin\frac{t}2 - \cos\frac{t}2\right)^2$ so we get $$A = 16$$

Is this result what you are getting?

$\endgroup$
  • $\begingroup$ exactly ! i did get this and with the same steps $\endgroup$ – Mather Mar 9 at 14:03
  • $\begingroup$ what is wrong with this answer i wrote up there $\endgroup$ – Mather Mar 9 at 14:03
  • $\begingroup$ i really cant find anything wrong with it $\endgroup$ – Mather Mar 9 at 14:03
  • $\begingroup$ wait i think i got the mistake the surface is $z(x,y)$ and i wrote something far beyond right . thanks for helping me !!! $\endgroup$ – Mather Mar 9 at 14:08
  • 1
    $\begingroup$ @i707107 Intuitively, the area element $dA$ of the cylinder is the product of the infinitesimal line element $d\gamma$ on the circle and the height of the cylinder over that line element, which is $z = \sqrt{4-x^2-y^2}$. Hence $$dA = \sqrt{4-x^2-y^2}\, d\gamma$$ Integrating this over $\gamma$ yields the surface area. $\endgroup$ – mechanodroid Mar 9 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.