Calculating Triple Integral using Cylindrical Coordinates

Question

I'm given $ E $ is located in $ x^2 + y^2 = (z-1)^2 $ and between $z = 0$ and $z=2$. I used level curves to graph this out, and as I see it is a circular cone. First, I set up my region, $$ E = \Big\{(r,\theta,z) | 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi, 0 \leq z \leq 2\Big\}.$$

I then set up my integral as $$ \int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{1} rdrdzd\theta $$

This resulted in $2\pi$, which was wrong. So then I changed the bounds of my $z$ by solving the equation for $z$, where $z = 1 \pm \sqrt{x^2+y^2} = 1 \pm r $ so that resulted in this integral: $$ \int_{0}^{2\pi}\int_{0}^{1}\int_{1-r}^{1+r} rdzdrd\theta $$ which resulted in $\frac{4}{3}\pi$, which was wrong as well. What am I doing wrong?

The only thing I can think of is to integrate on the equation $z = 1 + \sqrt{x^2+y^2}$ but I am not sure that is right either.

Answer 1

In case you didn't notice, the vertex of the cone equation is $(0,0,1)$ which is between the two planes. So the region has two "components" which touch only at that vertex, above and below it.

The first attempt

$$ \int_0^{2\pi} \int_0^2 \int_0^1 r\, dr\, dz\, d\theta \tag{1}$$

is the volume of all the points where $0 \leq z \leq 2$ and $r \leq 1$. This is a circular cylinder. As you noticed, the big issue is that it doesn't even use the equation of the cone.

In your next integral

$$ \int_0^{2\pi} \int_0^1 \int_{1-r}^{1+r} r\, dz\, dr\, d\theta \tag{2}$$

let's look at what the bounds say. $\theta$ can be in any direction. $r$, the distance from the $z$-axis, is never more than $1$. So far we have the same circular cylinder, but extending infinitely in the $+z$ and $-z$ directions. At any specific values of $\theta$ and $r$ (which also give specific values of $x$ and $y$), we then take the vertical line segment between $z=1-r$ and $z=1+r$ to be inside the region. That is, the segment is above the lower half-cone and below the upper half-cone.

So this integral actually gives the volume outside the (double) cone and inside the cylinder. One hint this wasn't the integral you wanted is that it doesn't use the bounding surfaces $z=0$ and $z=2$, but does use the bounding surface $r=1$. (Not directly, anyway. The $r=1$ does come from finding the intersection of the cone with each plane, but that's really a circle intersection and would only apply if we already know $z=0$ or $z=2$.)

But it turns out the region $E$ is exactly the cylinder from integral $(1)$, excluding the region outside the cone from integral $(2)$, so you do have a solution:

$$ \iiint_E dV = \int_0^{2\pi} \int_0^2 \int_0^1 r\, dr\, dz\, d\theta - \int_0^{2\pi} \int_0^1 \int_{1-r}^{1+r} r\, dz\, dr\, d\theta = 2\pi - \frac{4\pi}{3} = \frac{2\pi}{3} $$

Other ways to solve this:

As the answer by José Carlos Santos points out, making $r$ the innermost limit lets us vary $r$ from $0$ (the $z$-axis) to $|z-1|$ (the cone), giving a circular slice at every $z$ value, and using just the one integral

$$ \int_0^{2\pi}\int_0^2\int_0^{|z-1|}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta = \frac{2\pi}{3} $$

Or, staying with the $dz\, dr\, d\theta$ order, the region of $E$ at a given $r$ and $\theta$ ($x$ and $y$) is two separate line segments: between the $z=0$ plane and the lower half-cone, and between the $z=2$ plane and the upper half-cone, giving

$$ \int_0^{2\pi} \int_0^1 \left[ \int_0^{1-r} r\, dz + \int_{1+r}^2 r\, dz \right]\, dr\, d\theta = \frac{2\pi}{3} $$

Answer 2

Assuming that what you want is to compute the volume of that region, then you can do it in cylindrical coordinates. Then the region becomes $\require{cancel}r^2=(z-1)^2$, or $r=|z-1|$, and you can do:\begin{align}\int_0^{2\pi}\int_0^2\int_0^{|z-1|}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta&=\cancel2\pi\int_0^2\frac{(z-1)^2}{\cancel2}\,\mathrm dz\\&=\frac{2\pi}3.\end{align}