I'm trying to evaluate the following triple integral.
$ \int_{0}^{1} \int_{0}^{\sqrt{16-x^2}} \int_{0}^{\sqrt{16-x^2-y^2}} \text{dz dy dx} $
It looks to me like I'll probably need to convert it to cylindrical coordinates in order to evaluate it but I'm not sure how to
On
You could still carry out the triple integral as it is, without resorting to the cylindrical coordinates. It is actually easier.
After integration over $z$,
$$ I=\int_{0}^{1}dx \int_{0}^{\sqrt{16-x^2}} \sqrt{16-x^2-y^2} dy $$
The resulting $y$-integral can be performed with integration-by-parts as,
$$ \int_{0}^{a} \sqrt{a^2-y^2} dy = \frac{1}{2}\left[ y\sqrt{a^2-y^2}+a^2\sin^{-1}\left(\frac{y}{a}\right) \right]_{y=0}^{y=a}= \frac{a^2}{4}\pi $$
where $a^2=16-x^2$. In the end, the integral over $x$ becomes, $$ I=\frac{\pi}{4} \int_{0}^{1} (16-x^2)dx = \frac{47}{12}\pi $$
On
We have $$\int_{0}^{1} \int_{0}^{\sqrt{16-x^2}} \int_{0}^{\sqrt{16-x^2-y^2}} \mathrm d z \,\mathrm d y \,\mathrm d x=\int_{0}^{1} \int_{0}^{\sqrt{16-x^2}}\sqrt{16-x^2-y^2} \mathrm d y \,\mathrm d x.$$
This may be done directly. Set $\sqrt{16-x^2}=a,$ then the integral becomes $$\int_{0}^{1} \int_{0}^{a}\sqrt{a^2-y^2} \mathrm d y \,\mathrm d x.$$ Now make the usual substitution $y=a\sin\phi,$ which makes the inner integral elementary, so that the whole integral becomes $$\int_{0}^{1} \frac{πa^2}{4}\text{d} x=\frac π4 \int_{0}^{1} {16-x^2}\text{d}x,$$ which is obviously elementary.
If the bounds for $x$ were $0\le x\le 4$ then the region of integration was part of a sphere with radius $4$ centered at the origin in the first octant
The integral evaluates the volume of that region therefore it is $$ \frac {1}{8} \frac {4\pi (4^3)}{3}=\frac {32\pi }{3}$$
However with $0\le x \le 1$ we split the integral into two parts. Let $\alpha =\pi/2-\sin^{-1}(1/4)$, then $$I=I_1+I_2$$ $$I_1=\int _0^\alpha \int _0^{\sec(\theta ) } \sqrt {1-r^2}r dr d\theta$$
$$I_2=\int _\alpha ^{\pi/2} \int _0^4 \sqrt {1-r^2}rdrd\theta$$