Question

Find the area of the cap cut from the Hemisphere x2 + y2 + z2 = 2, by the cylinder x2 + y2 = 1 ('z' is greater than or equal to 0).
I need a perfect solution to check with my answer.
!!! Incorrect, Incomplete and spam answers for points are going to be reported, so answer only if u know this concept !!!​

Answer 1

Find the area of the cap cut from the Hemisphere x2 + y2 + z2 = 2, by the cylinder x2 + y2 = 1 ('z' is greater than or equal to 0).

I need a perfect solution to check with my answer.

!!! Incorrect, Incomplete and spam answers for points are going to be reported, so answer only if

Answer 2

Answer:

The required surface area is $2 \pi(2-\sqrt{2}) .$

Step-by-step explanation:

Given,

Hemisphere by the cylinder where $z$ is greater than equal to 0 .

We need to calculate the surface area of the cap of the hemisphere.

$\mathrm{K}=x^{2}+y^{2}+z^{2}=2$by the cylinder, $\mathrm{k}=x^{2}+y^{2}=1$

Now, let the surface be $S$ and it projects $x-y$ plane onto the circular disk $x^{2}+y^{2} \leq \mathrm{s}(\mathrm{R})$

We also see that $p=k$ is normal to$R$.

Now,

$&f(x, y, z)=x^{2}+y^{2}+z^{2}$

$&{Grad}(f)=2 x i+2 y j+2 z k$

$&|G r a d(f)|=2 \sqrt{2}$

Using the formula for the surface area we get,

Surface Area $=\iint_{R}|G r a d(f) /| G r a d(f) \cdot p \mid d A$

$\begin{aligned}&=\iint_{R} \sqrt{2} / 2 z d x d y \\&=\sqrt{2} \iint_{R} 1 / z d x d y\end{aligned}$

Now,

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

Therefore, $\mathrm{S}=\sqrt{2} \iint_{R} 1 / \sqrt{2-x^{2}-y^{2}} d x d y$

Now, converting it to polar co-ordinate:

Substituting $2-r 2=t^{2} ,$

$\Rightarrow-2 r \cdot d r=2 t \cdot d t$

$\mathrm{S}=\sqrt{2} \int_{q}^{p} \int_{b}^{a}-t . d t / t d( theta)$where, $\mathrm{a}=1$ and $\mathrm{b}=\sqrt{2} \\ \mathrm{p}=2 \pi$and $\mathrm{q}=0$

$\mathrm{S}=\sqrt{2} \int_{b}^{a} \sqrt{2}-1 d(theta) \quad$where, $\mathrm{a}=2 \pi ,\mathrm{b}=0$

$S=\sqrt{2}(\sqrt{2}-1) 2 \pi$

$S=2 \pi(2-\sqrt{2})$

Hence, the required surface area is $2 \pi(2-\sqrt{2}) .$