Finding the volume of an object (another)

Question

T is shaped like a quarter of a donut, and can be described in cylindrical coordinates by the inequality (r−2)^2 + z^2 ≤ 1. 0≤θ≤π/2. I am asked to find the volum of T. I think you can parameterize it like this: r = 2+rcos(θ), z = rsin(θ). But how do I set up the triple integral? This is really just a circle with radius 1 and center in (2,0).

Answer 1

By Pappus's Centroid Theorem we could calculate the volume in a simple way by

$$V=A\cdot \frac{\pi}2 R=\pi\cdot \pi=\pi^2$$

To proceed by direct integral calculation we have

$$(r-2)^2+z^2\le 1 \iff r^2-4r+4+z^2-1\le 0 $$

then since

$$r^2-4r+4+z^2-1= 0 \implies r=\frac{4\pm \sqrt{16-16+4-4z^2}}{2}=2\pm \sqrt{1-z^2}$$

we have that fo any $z\in[-1,1]$

$$2- \sqrt{1-z^2}\le r\le 2+\sqrt{1-z^2}$$

and then integral calculation in cylindrical coordinates we have

$$2\int_0^1 \frac{\pi}4 (r_{max}^2-r_{min}^2)dz=4\pi\int_0^1 \sqrt{1-z^2}dz=\pi^2$$