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

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

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?

• exactly ! i did get this and with the same steps – Mather Mar 9 at 14:03
• what is wrong with this answer i wrote up there – Mather Mar 9 at 14:03
• i really cant find anything wrong with it – Mather Mar 9 at 14:03
• wait i think i got the mistake the surface is $z(x,y)$ and i wrote something far beyond right . thanks for helping me !!! – Mather Mar 9 at 14:08
• @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. – mechanodroid Mar 9 at 14:21