Parameterize this surface by
r (u, v) = 9 cos(u) sin(v) i + 9 sin(u) sin(v) j + 9 cos(v) k
with 0 ≤ u ≤ 2π and 0 ≤ v ≤ arccos(4/9)
See the attached sketch to see how we arrive at the upper limit for v.
Take the normal vector to the surface to be
n = ∂r/∂u × ∂r/∂v
n = -81 cos(u) sin²(v) i - 81 sin(u) sin²(v) j - 81 sin(v) cos(v) k
which has magnitude
||n|| = 81 sin(v)
Then the surface area is
[tex]\displaystyle \iint_S \|\mathbf n\| \,\mathrm du\,\mathrm dv = 81 \int_0^{2\pi} \int_0^{\arccos(4/9)} \sin(v)\,\mathrm dv\,\mathrm du \\\\ = 162\pi \int_0^{\arccos(4/9)} \sin(v)\,\mathrm dv \\\\ = -162\pi (\cos\left(\arccos\left(\frac49\right) - \cos(0)\right) \\\\ = 162\pi \left(1 - \dfrac49\right) \\\\ = \boxed{90\pi}[/tex]
If you're not familiar with surface integrals, you can instead use what's sometimes called the projection method. Let
z = f(x, y) = √(81 - x ² - y ²)
(where we take the positive square root because we're looking at a part of the top half of the sphere)
Projecting the surface down onto the (x, y)-plane, we see that it casts a "shadow" of a disk with radius √(9² - 4²) = √65. (Use the Pythagorean theorem to solve for the missing side of the triangle in the sketch.)
Then the surface S considered above is hovering over the set in the (x, y)-plane,
D = {(x, y) : x ² + y ² ≤ 65}
The area is then
[tex]\displaystyle \iint_D \sqrt{1 + \left(\dfrac{\partial f}{\partial x}\right)^2 + \left(\dfrac{\partial f}{\partial y}\right)^2} \,\mathrm dx\,\mathrm dy[/tex]
It's easier to compute this integral in polar coordinates, so we take
x = r cos(t )
y = r sin(t )
dx dy = r dr dt
and the region D is given by the set
{(r, t ) : 0 ≤ r ≤ √65 and 0 ≤ t ≤ 2π}
Then the integral would be
[tex]\displaystyle \iint_D \sqrt{1 + \left(\dfrac{\partial f}{\partial x}\right)^2 + \left(\dfrac{\partial f}{\partial y}\right)^2} \,\mathrm dx\,\mathrm dy = \iint_D \sqrt{1 + \frac{x^2+y^2}{81-x^2-y^2}}\,\mathrm dx\,\mathrm dy \\\\ = \iint_D \sqrt{\frac{81}{81-x^2-y^2}}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^{2\pi} \int_0^{\sqrt{65}} r \sqrt{\frac{81}{81-r^2}}\,\mathrm dr\,\mathrm dt[/tex]
Substitute s = 81 - r ² and ds = -2r dr :
[tex]\displaystyle \int_0^{2\pi} \int_0^{\sqrt{65}} r \sqrt{\frac{81}{81-r^2}}\,\mathrm dr\,\mathrm dt = 2\pi \int_0^{\sqrt{65}} r \sqrt{\frac{81}{81-r^2}}\,\mathrm dr \\\\ = -9\pi \int_{81}^{16} \frac{\mathrm ds}{\sqrt s} \\\\ = 9\pi \int_{16}^{81} s^{-1/2}\,\mathrm ds \\\\ = 18\pi \left(\sqrt{81} - \sqrt{16}\right) \\\\ = \boxed{90\pi}[/tex]
