Find the volume inside the cone $z+2=\sqrt{x^2+y^2}$ between the planes $z=0$ and $z=1$.

I believe my limits with respect to $r$ are $0..2$ and $0..2\Pi$ with respect to $\theta$ (using the sketch in the xy plane). I think my limits with respect to $z$ are $0..r-2$. What are the correct limits of integration?

  • $\begingroup$ So, did you get a wrong answr w.r.t your assumed limits? $\endgroup$ – mrs Apr 16 '18 at 6:59
  • $\begingroup$ Yep. I used wolfram to compute the integration. $\endgroup$ – Stuart-James Burney Apr 16 '18 at 7:02
  • $\begingroup$ Why $r=0..2$? Don't you think it may be wrong? $\endgroup$ – mrs Apr 16 '18 at 7:04
  • $\begingroup$ I didnt think so. In the $xy$-plane, the equation is $x^2+y^2=4$. Isn't the maximum r can be 2 and the minimum 0? $\endgroup$ – Stuart-James Burney Apr 16 '18 at 7:09

Sorry for the delay! I am babysitting my naughty girl.

As you see, if you want to describe the triple integrals as you wanted so, we will get two sections (a cylinder inside and a part between cone and the cylinder). And we shoud illustrate the limits separately. So we have two district integrals.

Now lets change the limits as $\theta|_0^{\pi/2}$, $z|_0^1$ and assume $r$ as a function on $z$. It seems easier that the our first description. Then $$r=0..z+2$$

enter image description here Of course you multiply the final answer by $4$ regarding the symmetric of whole shape.


Everything starts with a good sketch:


The particular object that we have is called a frustum.

As our solid is symmetric with respect to rotation about the $z$-axis, you are correct that $0 \leq \theta \leq 2 \pi$.

We have two choices: bound $z$ in terms of $r$, or bound $r$ in terms of $z$.

If we fix a value for $z \in [0,1]$, then $r$ ranges from $0$ out to the edge of a cone at a value of...

$z + 2 = r$.

If we choose to bound $z$ in terms of $r$, then we will need to divide our integral into two regions---a central central cylinder (where the $z$ values range from $0$ to $1$), and an outer section where the $z$ values start at the surface of the cone and extend upwards to $z=1$.

We thus have two different ways to set up a triple integral for the volume, as well as geometric formula. Make sure that you get the same value for all of them!


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