# Triple Integral with Spherical Polar Coordinates Problem

I am having trouble understanding the whole concept of spherical polar coordinates, so any help is appreciated! The full question is:

Let D be the region bounded by the surface of the hemisphere $z = \sqrt{1 − x^2 − y^2}$ and the plane $z = 0$. Use spherical polar coordinates to find $$\iiint_{D}^{}z^3 \ dx \ dy \ dz$$

Answer: I have the answers, but I am unsure how/why they are what they are:

Spherical polar ranges; $0\leq r\leq 1 \$, $\ 0\leq \theta \leq \frac{\pi}2 \$, $\ 0 \leq\phi\leq 2\pi \$. Hence $$=\int_{0}^{2\pi} \int_{0}^{\frac{\pi}2} \int_{0}^{1}(r^3 \ cos^3\theta \ r^2 \ sin\theta\ )\ dr \ d\theta \ d\phi$$ $$=\frac{1}6 \frac{1}4 \int_{0}^{2\pi}d\phi = \frac{\pi}{12}$$

I am not concerned with the calculation of the integral, just how to get to that first line! Thank you!

Update:

$x=r sin \theta cos\phi \$, $y=rsin\theta sin\phi \$ ,$z=rcos\theta$ but I still don't know how i get the polar from these.

• Get the formulas for $x,y,z$ in spherical, and the volume element. Then plug in... – coffeemath Nov 16 '15 at 16:10
• @coffeemath I dont understand what you mean – jessicarevelo Nov 16 '15 at 16:20

If you square the equation for $D$ you get $z^2=1-x^2-y^2$ or $x^2+y^2+z^2=1$ You should recognize that as the unit sphere, which should tell you that the limits on $r$ are $0$ to $1$. The $\theta$ coordinate is like latitude on the earth except it is zero at the north pole and increases to $\pi$ at the sourth pole. The volume with $z \ge 0$ is the northern hemisphere, so $0 \le \theta \le \frac \pi 2$. $\phi$ is like longitude, but it runs from $0$ to $2 \pi$ and you want the whole range. That gives you the limits on the integral.
In the quantity to integrate, you substitute in $z=r \cos \theta$. The rest is the volume element in spherical coordinates.
• In many versions, $\theta$ is the angle measured downward from the north pole so $0 \le \theta \le \pi$ on the sphere. In the upper hemisphere the range of this $\theta$ is then from $0$ to $\pi/2$ as in the post. – coffeemath Nov 16 '15 at 16:32
• I'm more used to using $\theta$ and $\phi$ in the reverse sense, with $\phi$ going down from the north pole. [Also typically $\rho$ not $r$ for the radius (distance to origin). Anyway +1 on this which now conforms to setup in OP's question. – coffeemath Nov 16 '15 at 20:07