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Use a parameterization of the lower portion cut from the sphere $x^2 + y^2 + z^2 = 4$ by the cone $z = \sqrt{x^2 + y^2}$ to express the area of the surface as a double integral. Then evaluate the integral.


My Work

We can parameterise this surface using spherical coordinates:

$(x, y, z) = (\rho\cos(\theta)\sin(\phi), \rho\sin(\theta)\sin(\phi), \rho\cos(\phi)) \ \forall \ \rho > 0, 2\pi \ge \theta \ge0, \pi \ge \phi \ge 0$.

$x^2 + y^2 + z^2 = 4$

$\implies x^2 + y^2 + z^2 = 4$

$\therefore \rho =$ radius $= 2$

Our parameterisation for the sphere is $(x, y, z) = (2\cos(\theta)\sin(\phi), 2\sin(\theta)\sin(\phi), 2\cos(\phi)) = \mathbf{r}(\theta, \phi)$

We must now find the domain of the parameters for the region cut out from the sphere by the cone. To find the intersection between the cone and the sphere, we substitute the expressions for $x$, $y$, and $z$ into the equation for the cone, $z = \sqrt{x^2 + y^2}$:

$2\cos(\phi) = \sqrt{4\cos^2(\theta)\sin^2(\phi) + 4\sin^2(\theta)\sin^2(\phi)} = 2\sin(\phi)$

$\implies \cos(\phi) = \sin(\phi)$

$\therefore \phi = \dfrac{\pi}{4}$ since $\phi \in [0, \pi]$


Upon review, I noticed that the solutions specified something that I missed:

One limit for the domain of $\phi$ is $\dfrac{\pi}{4}$. The other limit is the value of $\phi$ at the bottom of the sphere $(0, 0, -2)$. Substitute $-2$ for $z$ in the parametric equation for $z$. Solve for the range $0 \le \phi \le \pi$.

Conceptually, I do not understand this part at all. I have reviewed my calculations and reasoning, but I still cannot understand anything about this section. Where does it comes from? How does it fit into the problem?

I do all of my reasoning analytically rather than by drawing pictures. From an analytical perspective, I am struggling to see what I missed in my reasoning and calculations.

I would greatly appreciate it if people could please take the time to clarify and elaborate on this section. I would also appreciate it if respondents would please be kind enough as to use analytical reasoning rather than relying on pictures/graphs.

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  • $\begingroup$ What did you take as the other limit for $\phi$ and why? $\endgroup$ – StackTD Apr 10 '17 at 12:56
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You found the value of $\phi$ corresponding to the intersection by doing:

To find the intersection between the cone and the sphere, we substitute the expressions for $x$, $y$, and $z$ into the equation for the cone, $z = \sqrt{x^2 + y^2}$:

$2\cos(\phi) = \sqrt{4\cos^2(\theta)\sin^2(\phi) + 4\sin^2(\theta)\sin^2(\phi)} = 2\sin(\phi)$

$\implies \cos(\phi) = \sin(\phi)$

$\therefore \phi = \dfrac{\pi}{4}$ since $\phi \in [0, \pi]$

but you actually want all the $\phi$-values corresponding to the part of the sphere under the cone; i.e. the values of $\phi$ not just corresponding to $z \color{red}{=} \sqrt{x^2+y^2}$, but to $z \color{blue}{\le} \sqrt{x^2+y^2}$; so: $$z \le \sqrt{x^2+y^2} \iff \cos \phi \le |\sin \phi| \quad\quad (0 \le \phi \le \pi)$$ Solving the inequality leads to $\phi \in \left[ \tfrac{\pi}{4} , \pi \right]$.

Is this what you were looking for? Feel free to comment if not.


Note that this agrees with the geometrical interpretation (you wanted to avoid) and also with:

One limit for the domain of $\phi$ is $\dfrac{\pi}{4}$. The other limit is the value of $\phi$ at the bottom of the sphere $(0, 0, -2)$. Substitute $-2$ for $z$ in the parametric equation for $z$. Solve for the range $0 \le \phi \le \pi$.

Since, again with $0 \le \phi \le \pi$: $$z=-2 \implies -2 = 2\cos \phi \iff \cos \phi = -1 \iff \phi = \pi$$

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  • $\begingroup$ This was exactly what I was looking for. Thank you for the excellent answer. $\endgroup$ – The Pointer Apr 10 '17 at 22:07
  • $\begingroup$ @ThePointer Alright, thanks and you're welcome! $\endgroup$ – StackTD Apr 11 '17 at 18:55

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