Triple integral cylindrical coordinates boundary

Question

I'm confused on the boundary of $r$ in this following volume question:
The volume between $z+2 = \sqrt{x^2 + y^2}$ and $z=0,z=1$.

I got the integral as
$$\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{z+2} r dr d\theta dz = \frac{19 \pi}{3}.$$
which is the correct answer. However, I don't understand the limits for $r$. Visually, I see that the minimum it can ever be (in this region) is actually $2$, which occurs on the $x-y$ plane. Why is it actually $0$?

Answer 1

The minimum value a radius can be $|r|$ is zero, and the maximum is governed by the $z+2$ term. From what you have posted, you wish to determine the volume of a solid shape, not a solid shape with a hole through the middle. If the minimum radius $|r|=2$, then you would have a hole with radius 2+2 through your object.

• 1: integrate w.r.t component $\theta$
• 2: integrate w.r.t component r, which inserts the boundary condition of the radius w.r.t component z
• 3: integrate w.r.t component z

w.r.t = with respect to

Edit: Reply to comment #1

Hmm, don't think I understand. If x^2+y^2=r^2, isn't the minimum of r equal to 2 ? This is the radius of the smallest circle cross-section for z≥0. And then the radius r only increases from there. (Basically, since the definition of r is the radius a circle for a fixed z, why isn't it at a minimum of 2 since the smallest circle has this radius on the x−y plane (z=0))?

Yes, the minimum radius, conditioned on the $f(z)=z+2$ boundary will give $r=2$. But, you are integrating the radius with respect to a condition not a value or an evaluated expression.

$\int^{f(z)}_0 ...dr$ gives a solid object bounded by $0$ and $f(z)$. When you substitute f(z) into the radius, you then automatically condition the inner radius on the x-y plane because the function $f(z)$ governs it. The evaluation of $\int^{1}_{0}dz$ sets the minimum radius as $r=2$ and the maximum at $r=3$. Essentially, it is not the evaluation of the radius integral that holds the minimum and maximum radii, it is the evaluation of the integral of z after the radius integration, which sets the radius values.

Edit 2: Further clarification and explicit formulation for question poster and other viewers

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

1. You are using radius and height-use cylindrical co-ordinates it makes the problem easier to deal with.

2. The revolution is 0 to $2\pi$, integrate w.r.t $\theta$ accordingly

3. The z component is bounded by $z_{Lower}$ and $z_{higher}$

4. The radius component is bounded by $r_{lower}$ and $r_{higher}(z)$

Initial formulation: $$\int_{0}^{2\pi}\int_{z_{lower}}^{z_{higher}}\int_{r_{lower}}^{r_{higher}(z)} r dr dz d\theta$$

Integrate and evaluate w.r.t $\theta$ $$2\pi\int_{z_{lower}}^{z_{higher}}\int_{r_{lower}}^{r_{higher}(z)} r dr dz$$

Now, the radius component is dependent upon component z, so integrate w.r.t r: $$2\pi\int_{z_{lower}}^{z_{higher}}\frac{r^2}{2} dz \Big|^{r_{higher}(z)}_{r_{lower}}$$

The lowest r can be is the value 0, and the highest it can be (unless bounded) is $\infty$. It is infact bounded by the z+2 condition. Therefore giving $$2\pi\int_{z_{lower}}^{z_{higher}}\frac{r^2}{2} dz \Big|^{z+2}_{0}= \pi\int_{z_{lower}}^{z_{higher}} ((z+2)^2-0^2) dz$$

$$ \pi\int_{z_{lower}}^{z_{higher}} z^2+4z^2+4 dz$$ $z_{lower}=0$ and $z_{higher}=1$

Evaluating to $\frac{19}{3}\pi$