Determining the bounds for a triple integral? I need help with a triple integration problem. I don't need help actually integrating this thing, I just need help with setting the actual integral(s) up. Specifically, I dont know how to determine what the bounds are.
We basically have a 3D donut. The problem says we can model the donut as a torus centered at the origin (0,0,0) with outer radius R=4 and inner radius r=2. The points (x,y,z) inside the torus are described by the following condition:
Equation Link
Here, c is the radius from the origin to the center of the torus tube (so I think it's 3), and a is the radius of the donut tube (so I think it should be 1), the cross section of the donut tube is a circle.
I need to calculate the volume of the donute after 2 cuts. The first cut happens parallel to the x axis at y = -3, and then parallel to the y axis at x = 1.
So far, I set up the following integral (in polar coordinates) that I think represents the volume of the whole donut:
Integral Eq
But I dont know how to subtract the cuts? I'm assuming I need two more integrals that have to be subtracted from the whole volume, but I cant set it up. Can anyone help? Is the volume I have correct so far?

You have a problem of order: The limits on the $z$ integral depend on $r$, and therefore it has to be done where $r$ is defined. In other words, the integral is $$\int_0^{2\pi}\int_2^4\int_{-\sqrt{a^2-(r-c)^2}}^{\sqrt{a^2-(r-c)^2}} rdzdrd\theta$$
And you are correct that $c = 3$ and $a = 1$
But as noted in the comments, this approach does not work well with the cuts. The problem is, the cuts are flat, which leads to complicated limits of integration for $r$ and/or $\theta$, which will undo the advantages of going to cylindrical representation ("polar" coordinates are in the plane - the 3D analogs are either "cylindrical", as you are using here, or "spherical").
Let's look at the cuts. The first is by the plane $y = -3$ (which is automatically parallel to the x-axis and to the z-axis). This plane is tangent to the "center ring" of the torus. The other plane is $x = 1$, which passes through the hole. Note that these two planes meet inside the torus. To set the up as an integral in $x, y, z$, you need to consider that for given values of $x$, there are in general two separate ranges of values for $y$ that need integrated over, and vice versa. The way to handle this is to break the torus up into quadrants and work each quadrant separately. Within each quadrant, there is only one range for $x$ or $y$ to be integrated. We can also do the same for $z$, limiting attention to $z \le 0$ and $z \ge 0$ separately. But in this case, the geometry is symmetric, so we will pick up the same volume on either side. Thus we can calculate only $z \ge 0$, then double it to get the whole volume.
The equation of the torus surface is $$z^2 +(\sqrt{x^2 + y^2} - 3)^2 = 1$$ So we can set up the integration over the entire first quadrant as
$$2\int_0^4\int_0^{\sqrt{16 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$
After cutting, for each quadrant we have:


*

*Quadrant 1: $0 \le x \le 1, 0 \le y$


$$2\int_0^1\int_0^{\sqrt{16 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$


*

*Quadrant 2: $x \le 0, 0 \le y$


$$2\int_{-4}^0\int_0^{\sqrt{16 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$


*

*Quadrant 3: $x \le 0, -3 \le y \le 0$


$$2\int_{-4}^0\int_{-3}^{\sqrt{4 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$


*

*Quadrant 4: $0 \le x \le 1, -3 \le y \le 0$


$$2\int_0^1\int_{-3}^{\sqrt{4 - x^2}}\int_0^{\sqrt{1-(\sqrt{x^2+y^2}-3)^2}} dzdydx$$
The $z$ integration is easy, but for the $y$ and $x$ integrations, you will need some trigonmetric substitutions.
