Calculate the surface integral $ \ \large \int_{D} xyz dS \ $, where the surface $D$ is that part of the sphere $x^2+y^2+z^2=4$, which located above the area $y \leq x, \ y \leq 0, \ 0 \leq x^2+y^2 \leq 4$.
Answer:
$x^2+y^2+z^2=4, \ 0 \leq x^2+y^2 \leq 4$ gives us the range of $z$ as $ \ 0 \leq z \leq 2$,
$y \leq x, \ y \leq 0, \ 0 \leq x^2+y^2 \leq 4$ gives us the range of $y \ $ as $-\sqrt 2 \leq y \leq 0$,
But what would be limit of $x$ ?
Help me
We have to divide the integral in two regions. In the third quadrant
y is from 0 to $\sqrt{4 - x^2} - x$ and x from $-\sqrt{2}$ to 0
In the 4th quadrant
y is from 0 to $\sqrt{4 - x^2}$ and x from 0 to 2