This problem is setup for spherical coordinates, so I would recommend you don't use the surface area formula you've written in terms of Cartesian coordinates.
Since $r=2$ is fixed, the infinitesimal area element in spherical coordinates is $dA=4\sin\theta d\theta d\phi$. The spherical cap is bounded by $0<\theta<\pi/4$ and $0<\phi<2\pi$. So try to evaluate the integral:
$$\int_0^{2\pi}\int_{0}^{\frac{\pi}{4}}4\sin\theta d\theta d\phi$$
I know this question was asked 5 years ago, but maybe I can be of help to future calc 3 students. I've spent 3 hours on this problem and finally figured it out...hopefully this helps!
Note: depending on your teacher's style, $u$ can be replaced with $r$ and $v$ can be replaced with $\theta$.
$$H(x,y,z)=x^2 \sqrt{5-4z},\quad z=1-x^2-y^2,\quad x≥0$$
$$\iint _S H(x,y,z)dσ= \iint_R H(u,v)|r_u×r_v | dudv$$
Convert to cylindrical:
$$x=u\cos v,\quad y=u\sin v,\quad x^2+y^2=u^2,\quad 0≤v≤2\pi$$
$$z=1-(x^2+y^2 )=1-u^2,\quad 0≤z≤1,\quad 0≤u≤1$$
$$H(u,v)=u^2\cos^2 v \sqrt{5-4(1-u^2 )}=u^2\cos^2 v\sqrt{4u^2+1}$$
Find determinant:
$$\vec{r}=u \cos v \hat{i} +u\sin v \hat{j}+(1-u^2 ) \hat{k}$$
$$\vec{r}_u=\cos v \hat{i}+ \sin v\hat{j}+(-2u)\hat{k}$$
$$\vec{r}_v=-u\sin v\hat{i}+u\cos v\hat{j}+0 \hat{k}$$
$$\vec{r}_u\times \vec{r}_v=(0-(-2u^2\cos v ))\hat{i}+(0-(-2u^2\sin v ))\hat{j}+(u \cos^2v+u \sin ^2 v )\hat{k}$$
$$=2u^2\cos v \hat{i}+2u^2\sin v \hat{j}+u\hat{k}$$
$$|\vec{r}_u\times \vec{r}_v|=\sqrt{4u^4\cos^2 v+4u^4\sin^2 v+u^2 }$$
$$=u\sqrt{4u^2+1}$$
Put it together:
$$H(u,v)|\vec{r}_u\times \vec{r}_v|=u^2 \cos^2 v \sqrt{4u^2+1} u\sqrt{4u^2+1}$$
$$=\cos^2 v (4u^5+u^3 )$$
$$\iint _RH(u,v)|\vec{r}_u\times \vec{r}_v| dudv$$
$$=\int_0^{2\pi}\int _0^1\cos^2v (4u^5+u^3 ) dudv$$
$$=\int_0^{2\pi}\cos^2v \left({4\over6} u^6+{{1\over4} u}^4 \right)|^{1}_{0} dv$$
$$=\int_0^{2\pi}{11\over12} \cos^2v dv$$
$$=\int_0^{2\pi}\left({11\over12}\right) \left({1\over2}\right)(1+\cos 2v )dv$$
$$=\left.\left({11\over24} v+{1\over2} \sin2v \right)\right|^{2\pi}_0$$
$$={11\pi\over12}$$
Best Answer
I think it is easiest to get your head around it if you break the region in two.
we have the hemisphere above the plane,
$S_1: z = \sqrt {4-x^2 - y^2}$
and the hemisphere below the plane
$S_2: z = -\sqrt {4-x^2 - y^2}$
$\iint f(x,y,z)\ dS_1 + \iint f(x,y,z) \ dS_2$
$\|dS_1\| = \|-z_x, -z_y, 1\| = \sqrt {4x^2 + 4y^2 + 1} \ dx\ dy$
$\|dS_2\| = \|z_x, z_y, -1\| = \sqrt {4x^2 + 4y^2 + 1} \ dx\ dy$
We want to keep the vectors pointing out, hence the sign change.
but $\|dS_1\| =\|dS_2\| $ it is the same for both hemispheres, so we can add the functions together.
$\iint [f(x,y,\sqrt {4-x^2 - y^2}) + f(x,y,-\sqrt {4-x^2 - y^2})]\sqrt {4-x^2 - y^2}\ dx\ dy$
and in this case
$f(x,y,z) + f(x,y,-z) = 0$
Update
converting to Spherical coordinates
$x = 2 \cos \theta\sin\phi\\ y = 2 \sin\theta\sin\phi\\ z = 2 \cos \phi$
$\|dS\| = \|(\frac {dx}{d\theta}, \frac {dy}{d\theta},\frac {dz}{d\theta})\times(\frac {dx}{d\phi}, \frac {dy}{d\phi},\frac {dz}{d\phi})\|\ d\theta\ d\phi = 4\sin\phi \ d\theta\ d\phi$
over the hemisphere:
$\int_0^{2\pi} \int_{0}^{\frac \pi2} (4cos^2\theta \sin^2\phi + 4\sin^2\theta\sin^2\phi)(2\cos \phi) (4\sin\phi) \ d\theta\ d\phi\\ \int_0^{2\pi} \int_{0}^{\frac {\pi}{2}} 32\sin^3\phi\cos \phi \ d\theta\ d\phi$
over the sphere
$\int_0^{2\pi} \int_{0}^{\pi} 32\sin^3\phi\cos \phi \ d\theta\ d\phi$