An alternative solution to the one I linked in the comments:
Let
z = f(x, y) = √(64 - x² - y²)
Then the integral of y² over the surface S is
[tex]\displaystyle \iint_S y^2 \, ds = \iint\limits_{x^2+y^2\le16} y^2 \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy[/tex]
We have
∂f/∂x = -x/√(64 - x² - y²)
∂f/∂y = -y/√(64 - x² - y²)
so that the integrand simplifies to
[tex]\displaystyle \iint_S y^2 \, ds = \iint\limits_{x^2+y^2\le16} y^2 \sqrt{1 + \frac{x^2}{64-x^2-y^2} + \frac{y^2}{64-x^2-y^2}} \, dx \, dy[/tex]
[tex]\displaystyle \iint_S y^2 \, ds = \iint\limits_{x^2+y^2\le16} y^2 \sqrt{\frac{64}{64-x^2-y^2}} \, dx \, dy[/tex]
[tex]\displaystyle \iint_S y^2 \, ds = 8 \iint\limits_{x^2+y^2\le16} \frac{y^2}{\sqrt{64-x^2-y^2}} \, dx \, dy[/tex]
To compute the remaining integral, convert to polar coordinates using
x = r cos(θ)
y = r sin(θ)
dx dy = r dr dθ
The region x² + y² ≤ 16 is given by the set
D = {(r, θ) : 0 ≤ r ≤ 4 and 0 ≤ θ ≤ 2π}
Then the integral becomes
[tex]\displaystyle \iint_S y^2 \, ds = 8 \int_0^{2\pi} \int_0^4 \frac{(r\sin(\theta))^2}{\sqrt{64-r^2}} (r \, dr \, d\theta)[/tex]
[tex]\displaystyle \iint_S y^2 \, ds = 8 \int_0^{2\pi} \int_0^4 \frac{r^3 \sin^2(\theta)}{\sqrt{64-r^2}} \, dr \, d\theta[/tex]
[tex]\displaystyle \iint_S y^2 \, ds = 8 \left(\int_0^{2\pi} \sin^2(\theta) \, d\theta\right) \left(\int_0^4 \frac{r^3}{\sqrt{64-r^2}} \, dr\right)[/tex]
Now,
[tex]\displaystyle \int_0^{2\pi} \sin^2(\theta) \, d\theta = \frac12 \int_0^{2\pi} (1-\cos(2\theta)) \, d\theta = \pi[/tex]
and by substituting u = 64 - r²,
[tex]\displaystyle \int_0^4 \frac{r^3}{\sqrt{64-r^2}} \, dr = -\frac12 \int_{64}^{48} \frac{64-u}{\sqrt u} \, du[/tex]
[tex]\displaystyle \int_0^4 \frac{r^3}{\sqrt{64-r^2}} \, dr = \frac12 \int_{48}^{64} \left(64u^{-\frac12} - u^{\frac12}\right) \, du[/tex]
[tex]\displaystyle \int_0^4 \frac{r^3}{\sqrt{64-r^2}} \, dr = \frac{1024}3 - 192\sqrt{3}[/tex]
So, the surface integral has a value of
[tex]\displaystyle \iint_S y^2 \, ds = \boxed{\left(\frac{8192}3 - 1536 \sqrt3\right) \pi}[/tex]
