Question

) Consider the circle of radius 2 centered on the x-axis at the point (5, 0). If you rotate this circle around the y-axis, you get a torus (a.k.a a doughnut). Find the surface area of the torus using an integral.

Answer 1

236.87 Square Units

Step-By-Step Explanation

Consider a torus generated by the rotation of the circle about the y-axis given below.

[TeX] (x-R)^2 + y^2 = r^2, \quad R > r > 0 [/TeX]

The perimeter of its cross section is 2\pi r. When it involves the y-axis:

Its total surface area:

[TeX]S=\int_{0}^{2\pi R}2\pi r d\alpha=4\pi^2 rR[/TeX]

For a circle radius 2 centered on the x-axis at the point (5, 0).

r=2. x=5, y=0

[TeX] (x-R)^2 + y^2 = r^2[/TeX]

[TeX] (5-R)^2 + 0^2 = 2^2[/TeX]

[TeX] (5-R)^2=2^2[/TeX]

5-R=2

R=5-2=3

Therefore, the total Surface Area is:

[TeX]= 4\pi^2 rR \\=4*2*3 \pi^2 \\=24 \pi^2=236.87 [/TeX]