question archive Find the surface area of the torus shown below, with radii r and R,

Find the surface area of the torus shown below, with radii r and R,

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Find the surface area of the torus shown below, with radii r and R,

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Let's consider a cross-section of the torus along its axis.  We see 
two circles of radius r, their centers at (-R,0) and (R,0).  Let's 
take an element of arc length, ds = r.d\theta , on the right-hand circle 
with \theta defined as the angle measured at the center of the small 
circle, from the positive x direction to the element under 
consideration.

This arc element sweeps out an area equal to ds times the 
circumference of the circle it sweeps out as the circle is rotated to 
form the torus. 

The radius of that circle is R + r\cos\theta.  Thus,
 
the circumference is 2\pi(R + r\cos\theta), 

and the surface area swept out is

dA = 2\pi(R+r\cos\theta).r.d\theta \ \ \ \ \ 0\leq\theta\leq2\pi

Therefore, Surface Area is

A = \int dA = \int_0^{2\pi}2\pi(R+r\cos\theta).r.d\theta \\ \\ \implies A = 2\pi r\bigg(\int_0^{2\pi}Rd\theta + \int_0^{2\pi}r\cos\theta d\theta\bigg)\\ \\ \implies A = 2\pi r(2\pi R + r(\sin\theta)|_0^{2\pi}) = 4\pi^2rR

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