Question

The region bounded by \((x-2)^{2}+y^{2}=1\) is revolved about the \(y\) -axis to form a torus. Find the surface area of the torus.

Short answer

The surface area of the torus is \(8π^2\)

Step-by-step solution

Derive the function

We start by deriving \(y = \sqrt{1 - (x-2)^2}\). The derivative is \(y' = -(x-2)\frac{1}{\sqrt{1 - (x-2)^2}}\)

Compute the magnitude function

We next calculate the magnitude function \(\sqrt{1+[y']^2}\). This results in \(\sqrt{1+\frac{(x-2)^2}{1 - (x-2)^2}} = \frac{1}{\sqrt{1 - (x-2)^2}}\)

Define the limits of the integral

The bounds of the integral are set by the domain of the original function, which is from 1 to 3.

Integrate to find the surface area

We compute the integral \(A = 2π\int_{1}^{3}(x)(-x+2)\frac{1}{\sqrt{1 - (x-2)^2}} dx\)

Simplify the integral

We can simplify the integral by noticing the symmetry about \(x=2\), and rewrite it as \(A = 4π(2\int_{1}^{3}dx-x\int_{1}^{2}\frac{1}{\sqrt{1 - (x-2)^2}} dx)\)

Calculate the value of the integral

The integral is then calculated as \(A = 4π(2(3-1)-2\int_{1}^{2}dx + 2\int_{1}^{2} dx)\) which eventually simplifies to \(A = 8π^2\)