Evaluate the surface integral of scalar function $\int_{S} (x + y + z )dS$ where $S$ is the boundary of the unit ball $B$ Evaluate the surface integral of scalar function $\int_{S} (x + y + z )dS$ where $S$ is the boundary of the unit ball $B$
my attempt
$S = \int_{S} (x + y + z) dS$
Unit ball $s: x^2 + y^2 + z^2 = 1 \to z^2 = 1 - x^2 - y^2$
$2z \frac{dz}{dx} = -2x$
$\frac{dz}{dx} = -x/z$
$2z\frac{dz}{dy} = -2y$
$\frac{dz}{dy} = -y/z$
since we know that
$\sqrt{1 + \left(\frac{dz}{dx} \right)^2 + \left( \frac{dz}{dy}\right)^2}dx dy = \sqrt{\frac{x^2 + y^2 + z^2}{z^2}}dxdy$
$ds = \frac{1}{2}dx dy$ (since $x^2 + y^2 + z^2 = 1$)
now set $z = 0$, then the limits are
$x$ goes to $-1$ and $1$
and $y$ goes to $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$m then
$$S = \int \int (x + y + z) \frac{1}{2}dxdy = \int \int \left( \frac{x}{\sqrt{1-x^2-y^2}} + \frac{y}{\sqrt{1-x^2-y^2}} + 1\right)dxdy $$
now set $z = 0$, then the limits are
$x$ goes to $-1$ and $1$
and $y$ goes to $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$m then
$$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left( \frac{x}{\sqrt{1-(x^2+y^2)}} + \frac{y}{\sqrt{1-(x^2+y^2)}} + 1\right)dxdy $$
since we know that $\int_{-a}^{a} f(x)dx = 0$ when $f(x)$ is a odd number
$$S = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} [0 + 0 + 1]dxdy = \pi$$
would this be correct?
 A: By linearity,
$$ \int_{S} (x + y + z) dS=\int_{S} xdS+\int_{S} ydS+\int_{S} zdS$$
Let $S_{x^+}$ the boundary of the unit ball in the half-plane $x\geq 0$ and define in a similar way $S_{x^-}$. Then
$$\int_{S} xdS=\int_{S_{x^+}} xdS+\int_{S_{x^-}} xdS=\int_{S_{x^+}} xdS+\int_{S_{X^+}} (-X)dS=0.$$
where in the second integral we let $X=-x$. In a similar way we prove that $\int_{S} ydS=\int_{S} zdS=0$, and we may conclude that the given integral is zero.
More generally, if $f$ is a integrable over the boundary of a ball $S$ (centered at the origin) and $f(-x,-y,-z)=-f(x,y,z)$ for all $(x,y,z)\in S$, then $\int_Sf(x,y,z)\,dS=0$.
P.S. In your work.
$$\sqrt{1 + \left(\frac{dz}{dx} \right)^2 + \left( \frac{dz}{dy}\right)^2}dx dy = \sqrt{\frac{x^2 + y^2 + z^2}{z^2}}dxdy=\frac{dxdy}{|z|}.$$ Therefore, since $z/|z|=1$ when $z>0$ and $z/|z|=-1$ when $z<0$, we have that
$$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left( \frac{x}{\sqrt{1-(x^2+y^2)}} + \frac{y}{\sqrt{1-(x^2+y^2)}} + 1\right)dxdy\\+\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left( \frac{x}{\sqrt{1-(x^2+y^2)}} + \frac{y}{\sqrt{1-(x^2+y^2)}} - 1\right)dxdy=\pi-\pi=0$$
