I'm not sure how to approach the problem, though I know it needs to be broken up into two integrals before it can be evaluated based on the way the answer input is set up. I do not know how to start visualizing what is going on correctly though, so any help on that front would be greatly appreciated.

Problem: $$\int_0^2 \int_0^\sqrt{4 - x^2} \int_0^\sqrt{16 - x^2 - y^2} \sqrt{x^2 + y^2} \, dz \, dy \, dx$$

• Did you try using the formulas for converting form Cartesian coordinates to spherical coordinates? If so, can we see your work? Mar 29, 2016 at 23:59

Hint: If $x \in [0, 2]$, then: $$y \leq \sqrt{4 - x^2} \iff y^2 \leq 4 - x^2 \iff x^2 + y^2 \leq 4 \iff r^2 \leq 4 \iff r \leq 2$$ Note that in the $xy$-plane, $\{(x, y) \mid x \in [0, 2] \text{ and } y \in [0, \sqrt{4 - x^2}\}$ represents the Quadrant I section of a circle centred at the origin with radius $2$ (that is, it is the top-right quarter circle). So what should $\theta$ range between?

When switching from Cylindrical to Spherical, imagine taking a cross-section by slicing the region at an arbitrary angle using the half-plane $\theta$ to obtain an $rz$-plane. In this cross-section, we want: $$S = \{(r, z) \mid r \in [0, 2] \text{ and } z \in [0, \sqrt{16 - r^2}]\}$$ That is, we are taking a quarter-circle of radius $4$ and making a vertical slice at $r = 2$ and keeping the left piece. Now because $(0, 4), (2, 0) \in S$, we are forced to let $\phi \in [0, \pi/2]$. As for $\rho$, it's a bit tricker as we transition from the arc $z = \sqrt{16 - r^2}$ to the line $r = 2$, which intersect at the threshold $(2, \sqrt{12})$. Notice that at this threshold, we have $\rho = 4$, so: $$2 = r = \rho\sin\phi = 4\sin\phi \implies \phi = \frac{\pi}{6}$$ Hence, for $\phi \in [0, \pi/6]$, the outermost boundary for $\rho$ is: $$z = \sqrt{16 - r^2} \iff r^2 + z^2 = 4^2 \iff \rho = 4$$ while for $\phi \in [\pi/6, \pi/2]$, the outermost boundary for $\rho$ is: $$r = 2 \iff \rho\sin\phi = 2 \iff \rho = 2\csc\phi$$

Putting it all together, we obtain: \begin{align*} &\int_0^2 \int_0^\sqrt{4 - x^2} \int_0^\sqrt{16 - x^2 - y^2} \sqrt{x^2 + y^2} \, dz \, dy \, dx \\ &= \int_0^{\pi/2} \int_0^2 \int_0^\sqrt{16 - r^2} r (r \, dz \, dr \, d\theta) \\ &= \int_0^{\pi/2} \int_0^{\pi/2} \int_0^{\min\{4, 2\csc\phi\}} (\rho \sin \phi)(\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta) \\ &= \int_0^{\pi/2} \left[ \int_0^{\pi/6} \int_0^{4} \rho^3 \sin^2 \phi \, d\rho \, d\phi + \int_{\pi/6}^{\pi/2} \int_0^{2\csc\phi} \rho^3 \sin^2 \phi \, d\rho \, d\phi \right] d\theta \end{align*}

• between 0 and pi/2. But I don't get where the need for a second integral is coming from. Could you possibly help me understand that? Mar 30, 2016 at 1:15
• Why shouldnt $r\leq 4$? For, you can apply the same reasoning to $z\leq \sqrt{16-x^2-y^2}$ with the substitution in spherical coordinates, $r=\sqrt{x^2+y^2+z^2}$. Mar 30, 2016 at 1:36
• I agree with the region for cylindrical coordinates, but I think the region for spherical coordinates is too large. For example, if $\theta=\phi=0$ and $\rho=4$ (which is on the boundary of the given region), then $x=4$, which is far outside the original region of integration. Mar 30, 2016 at 13:33
• @DavidK I think you meant $\phi = \pi/2$, but I get what you mean. Thanks, I updated my answer. Mar 30, 2016 at 16:56
• Yes, I meant a point whose "standard" spherical coordinates have $\phi=\pi/2$. I spend too much time working with geodetic coordinates on the Earth, where $\pi/2$ is the latitude of the north pole (in radians). Anyway, the answer looks fine now. Mar 30, 2016 at 19:20

Consider an integral of this form: $$\int_0^4 \int_0^{\sqrt{16-x^2}} \int_0^{\sqrt{16-x^2-y^2}} f(x,y,z) \, dz\, dy\, dx. \tag1$$ The region of integration of $f(x,y,z)$ in this integral is the portion of a sphere of radius $4$ around the origin for which $x>0$, $y>0$, and $z>0$. \tag2

Next, consider an integral of the form $$\int_0^2 \int_0^{\sqrt{4-x^2}} \int_0^4 f(x,y,z) \, dz\, dy\, dx.$$ The region of integration of $f(x,y,z)$ in this integral is the portion of a cylinder of radius $2$ and height $4$ around the $z$-axis for which $x>0$ and $y>0$.

Compare your region of integration. It is smaller than the sphere, because it eliminates all the points where $x>2$, and when $x\leq 2$ it eliminates all the points where $y > \sqrt{4-x^2}$. That is, it eliminates everything outside a cylinder of radius $2$ around the $z$-axis. But your region of integration is smaller than the cylinder of radius $2$ and height $4$, because when $x>0$ or $y>0$, the upper bound of $z$ is $\sqrt{16-x^2-y^2} < 4$.

What you actually have is the intersection of the two regions of integration in formulas $(1)$ and $(2)$. If you take a ray straight out from the origin, depending on the direction in which you go, you're either going to hit the surface of the sphere first, or you're going to hit the surface of the cylinder first, and when either of those happens the ray passes outside the region of integration, so that point is at your maximum value of $\rho$ in that direction.

The reason you need two integrals is to have one for all the directions where $\rho$ is limited by the surface of the sphere, and another for the directions where $\rho$ is limited by the surface of the cylinder.