# Finding the limit of surface integral

I have found this problem in my textbook.

Evaluate $$\iint_S F\cdot n ds$$ where $$F = yz\hat i+xz\hat j+xy\hat k$$ and S is that part of the surface of the sphere $$x^2+y^2+z^2 = 1$$ which lies in first octant.

I have tried to solve this problem using $x = rcos\theta \cdot sin\phi$ and $y = rsin\theta \cdot sin\phi$ . But i'm stuck in determining the limit. In my text book they used $\phi : 0\to\pi/2$ and $\theta : 0\to\pi/2$ . But i couldn't understand it. Can anyone please explain it? Thanks in advance.

The reason why $\phi$, the azimuthal angle only goes from $0$ to $\frac{\pi}{2}$ is because we only take a quarter-turn along the circle in the $xy$-plane. Since there are $2\pi$ radians in a circle and we are only taking a quarter turn, this reduces to $\frac{1}{4}2\pi = \frac{\pi}{2}$. As for $\theta$, the polar angle, the range of this from the $+z$-axis to the $-$z-axis is $\pi$ radians. Because we're only interested in the first octant, or the part where $z>0$, then we only go half the total distance (we arc from the $+z$-axis down to the $xy$-plane), and so we only $\frac{\pi}{2}$ radians for $\theta$.
• No, it means $1/8$th of the sphere. Imagine first taking the top half of the sphere. Now with the top half, you cut it up into quarters and take only one of the quarters. One of these last pieces represents the first octant and is only $1/8$th the whole object. – Mnifldz Jan 13 '17 at 3:54