In $\mathbb R^3$, let $C$ be the circle in the $xy$-plane with radius $2$ and the origin as the center, i.e.,

$$C= \Big\{ \big(x,y,z\big) \in \mathbb R^3 \mid x^2+y^2=4, \ z=0\Big\}.$$

Let $\Omega$ consist of all points $(x,y,z) \in \mathbb R^3$ whose distance to $C$ is at most $1$. Compute $$\int_\Omega \left|\,x\,\right|\,dx\,dy\,dz$$

So, with the help of erichan (see below), I now know that the volume is a solid torus. But I am having trouble setting up the integration bounds. As erichan had suggested, we consider a union of unit-spheres, all centered on points of the radius-$2$ circle. Using spherical coordinates, I have this integral set up:

$$\int_0^\pi\int_0^{2\pi}\int_0^1 \Big(\big|\,\left(r\cos \theta+2\right)\sin\phi \,\big| \,r^2 \sin\phi \Big) \, dr\, d\theta \,d\phi,$$

Where I parameterized the solid torus as: $$ \begin{aligned} x &= (r\cos \theta +2)\sin \phi \\ y &= (r\sin \theta +2)\sin\phi \\ z &= r\cos \phi \end{aligned} $$ with Jacobian as $r^2\sin \phi$.

Is my setup ok? I shifted $x$ and $y$ by two units. I'm not so sure about the parametrization of $z$ (should I leave it as it is normally?). And should I change the Jacobian factor?

I welcome any answers to this problem – I had previously requested just hints. I am wondering whether there is a simpler way to compute this integral, using symmetry of the torus.


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    $\begingroup$ I added in some details - your parametrization is a little off as is, so I included the proper parametrization. $\endgroup$ – izœc Aug 3 '15 at 5:40

So, you know it is a torus: then, your parametrization should be $$ \begin{cases} x = \big(\, 2 + r \cos \theta \, \big) \cos \phi \\ y = \big(\, 2 + r \cos \theta \, \big) \sin \phi \\ z = r \sin \theta \end{cases} $$ where $r \in [0,1]$ and $\theta, \phi \in [0, 2 \pi)$, and with jacobian $$ \big|\,J\,\big| = \left| \begin{vmatrix} \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial r} \\ % \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial r} \\ % \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial r} \\ \end{vmatrix} \right| = \cdots = r \,\left( \,2 + r \cos \theta\,\right) . $$ Thus your integral becomes: $$ \int_0 ^1 \int_0 ^{2 \pi} \int_0 ^{2 \pi} \big| \left(\,2 + r \cos \theta\,\right) \cos \phi \big| \, r \, ( \,2 + r \cos \theta\,) \, d \phi \, d \theta \, dr $$ $$ = \int_0 ^1 \int_0 ^{2 \pi} r\, \left( \,2 + r \cos \theta \,\right)^2 \, d \theta \, d r \int_0 ^{2 \pi} \big| \cos \phi \big| \, d \phi $$ which simplifies to $$ \left[ \int_0 ^{2 \pi} \big| \cos \phi \big| \, d \phi \right] \cdot \left[ 4 \int_0 ^1 r \, dr \int_0 ^{2 \pi} d \theta + 4 \int_0 ^1 r^2 \, dr \int_0 ^{2 \pi} \cos \theta \, d \theta + \int_0 ^1 r^3 \, dr \int_0 ^{2 \pi} \cos ^2 \theta \, d \theta \right] $$ which, while tedious, is rather trivial to compute.

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    $\begingroup$ @Vlad thanks for compensating for my clumsy fingers - it looks much better now. $\endgroup$ – izœc Aug 3 '15 at 6:01
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    $\begingroup$ @LebronJames To understand the torus parametrization, think of it this way: You have a circle flat in the plane, which you are surrounding like a tube with another circle. $r$ is the radius of this second circle, so we want to expand $r$ from $0$ to $1$ here. The two angles each "parametrize" one of the circles - one angle tells you where you are on the flat circle in the plane, while at ever point on that circle, the other angle will parametrize another circle around it. $\endgroup$ – izœc Aug 3 '15 at 7:00
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    $\begingroup$ @LebronJames This picture kind of explains what I am trying to say: math.rutgers.edu/~greenfie/currentcourses/math291/gifstuff/… $\endgroup$ – izœc Aug 3 '15 at 7:00
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    $\begingroup$ @LebronJames If you have done a very little bit of differential geometry in your multivariable calculus, you can understand this really, really well in terms of the moving frame / coordinate basis, using the tangent $\mathbf{T}$, normal $\mathbf{N}$, and binormal $\mathbf{B}$ vectors as your basis! But anyhow, think of it this way - in $z$, you want a circle, so you need $r$ and an angle, and in the $x$ and $y$, you need to do a circle with the other angle (NOT the $z$ angle!), but multiply by something that will give a circle in the z direction with that angle - so $\phi$ runs around the... $\endgroup$ – izœc Aug 3 '15 at 7:04
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    $\begingroup$ @LebronJames plane, while at every fixed $\phi$, we can run a circle about that along the $z$ axis using its angle $\theta$, and filling it in using $r$. Lastly (sorry to totally overwhelm with the length of this!), to do the integral of $|\cos|$ is actually very straightforward: break $[0, 2\pi)$ up into the sections where $\cos$ is positive and do $\int \cos$ on that, and add it to the integral $\int (-\cos)$ where it is taken over the part of the domain where $\cos$ is negative (and thus $- \cos$ is positive. $\endgroup$ – izœc Aug 3 '15 at 7:08

Since you only want hints...

For each point on the circle of radius $2$ consider a solid sphere of radius $1$. Now consider the union of all these spheres. What do you get?

  • $\begingroup$ That's so cool, @erichan... $\endgroup$ – user258501 Aug 3 '15 at 2:07
  • $\begingroup$ I seem to get a bunch of radius 1 disks, but all going in a circular direction to cover the radius-2 circle. So is it like a cylinder, but going in a circle? $\endgroup$ – user258501 Aug 3 '15 at 2:08
  • $\begingroup$ They are spheres, not disks. $\endgroup$ – zzchan Aug 3 '15 at 2:13
  • $\begingroup$ You get what is called a torus $\endgroup$ – zzchan Aug 3 '15 at 2:43
  • $\begingroup$ Oh, gosh, that was my first guess but didn't want to say it, since this is preparation for an advanced calc exam, not topology. So, I kept trying to force something cylindrical to come out ... $\endgroup$ – user258501 Aug 3 '15 at 2:44

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