Answer:
The question is not correct.
The correct question is sphere x2 +y2 + z2 = 36, f(x,y,z) = x2+y2 for z ≥ 0
Step-by-step explanation:
The given function is [tex]f(x,y,z) = x^2 + y^2, $[/tex] where S is the hemisphere, [tex]$ x^2 +y^2+ z^2 =36\ for\ z \geq 0$[/tex].
To evaluate the surface integral [tex]$ \int \underset{S} \int f dS $[/tex]
Now using the polar coordinates, we get
x = r cos θ, y = r sin θ for 0 ≤ θ ≤ 2π and x² + y² = r²
Now given
x²+y²+z=36 ......(i)
Put z = 0 in equation (i), we get
⇒ r² = 36, r = 6
Now writing the limits,
0 ≤ r ≤ 6, 0 ≤ θ ≤ 2π and f(x,y,z) = x²+y²=r²
Hence, surface integral
[tex]$ \underset{0}{\overset{2\pi} \int} \underset{0}{\overset{6} \int}r (r^2) dr d \theta $[/tex]
[tex]$ =\underset{0}{\overset{2\pi} \int} [\frac{r^4}{4}]_0 ^6 d \theta $[/tex]
[tex]$= 2 \pi ( \frac{6^4}{4}) = 648 \pi$[/tex]
