Let $D$ be the region in $\mathbb{R}^3$ below $z=-\sqrt{x^2 + y^2}$ and above $z=-\sqrt{4-x^2 -y^2}$.

Rewrite \begin{align*}\iiint \limits_D z^2 dV\end{align*} using Spherical Coordinates.

I rewrote the integral in spherical coordinates, and now know that it is as follows: \begin{align*}\iiint \limits_D z^2 dV &= \int\limits_{\theta_0}^{\theta_1}\int \limits_{\phi_1}^{\phi_2}\int\limits_{\rho_1}^{\rho_2}(\rho^4\cos^2{\phi}sin{\phi})d\rho d\phi d\theta \end{align*}

My problem, however, is finding the limits for $\rho, \phi$ and $\theta$ in the triple integral.

I am having some trouble finding these limits. Can somebody please assist me, as I honestly have no idea how to go about doing so for this question.

EDIT: $\textbf{I have to draw a sketch of the region $D$ in the given question (but I am struggling to do so)}$ and use that to find the limits.

  • $\begingroup$ The exact question is given as follows: Let $D$ be the region in $\mathbb{R}^3$ above the hemisphere $z=-\sqrt{4-x^2 -y^2}$ and below the cone $z=- \sqrt{x^2 + y^2}$. Evaluate the integral $\iiint\limits_D z^2 dV$ by using spherical coordinates. $\endgroup$ – user860374 Sep 19 '14 at 11:26

The lower cone begins at $\phi=\dfrac{3\pi}{4}$ and runs all the way down to $\phi=\pi$. If the starting point for $\phi$ isn't obvious, then you can find it as follows

Let $x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi, z=r\cos\theta$. Subbing this into the cone equation, squaring and rearranging you obtain

$$r^2\cos^2\theta = r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi\iff \tan^2\phi=1$$

Since we're in the lower octants, this means $\tan\phi=-1$, i.e. $\phi=\dfrac{3\pi}{4}$.

Obviously, $0\le \theta\le 2\pi$.

$\rho$ will fun from the origin to the hemisphere. $0 \le \rho\le 2$.

  • $\begingroup$ Thank you! This makes a lot of sense, algebraically :). But would it be possible for you to please send a rough sketch of what it would look like graphically, as I am still struggling a bit to picture what it would look like. $\endgroup$ – user860374 Sep 19 '14 at 12:41
  • $\begingroup$ My MS Paint skills would confuse you even more. It would look like this, but upside down tutorial.math.lamar.edu/Classes/CalcIII/TISphericalCoords_files/… $\endgroup$ – David Peterson Sep 19 '14 at 12:49
  • $\begingroup$ Something like this? imgur.com/OL2aWgR Sorry, imgur was the only photo sharing I could access right now $\endgroup$ – user860374 Sep 19 '14 at 13:05
  • $\begingroup$ Almost. The point of the cone should be below the xy plane $\endgroup$ – David Peterson Sep 19 '14 at 13:08
  • $\begingroup$ So the axes must be moved to the top op the cone? :). $\endgroup$ – user860374 Sep 19 '14 at 13:10

Consider $z=-r$ and $z=-\sqrt{4-r^2}$. You have that if $r>0$, so $-r=-\sqrt{4-r^2}\iff r^2=4-r^2\iff r=\sqrt 2$. Then, $$D=\left\{(r\cos\theta,r\sin\theta, z)\mid \theta\in[0,2\pi[, r\in[0,\sqrt 2], z\in[0,-\sqrt 2]\right\}$$$$\cup\left\{(r\cos\theta\sin\varphi, r\sin\theta\sin\varphi, r\cos\varphi)\ \big|\ r\in[0,2], \theta\in[0,2\pi[,\varphi\in\left[\frac{3\pi}{4},\pi\right]\right\}$$


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