Answer
(a) We show that
${\rm{Area}}\left( T \right) = 4\pi \mathop \smallint \limits_{b - a}^{b + a} \dfrac{{ay}}{{\sqrt {{a^2} - {{\left( {b - y} \right)}^2}} }}{\rm{d}}y$
(b) We show that
${\rm{Area}}\left( T \right) = 4{\pi ^2}ab$
Work Step by Step
(a) The circle in the $yz$-plane of radius $a$ centered at $\left( {0,b,0} \right)$ has equation
${\left( {y - b} \right)^2} + {z^2} = {a^2}$
So, we obtain $z = g\left( y \right) = \pm \sqrt {{a^2} - {{\left( {y - b} \right)}^2}} $.
First, we consider the upper half of the torus, so we use the graph $g\left( y \right) = \sqrt {{a^2} - {{\left( {y - b} \right)}^2}} $.
So,
$g'\left( y \right) = \dfrac{{ - \left( {y - b} \right)}}{{\sqrt {{a^2} - {{\left( {y - b} \right)}^2}} }}$
By Eq. (14) we compute the surface area of the upper half of the torus:
${\rm{UpperArea}}\left( T \right) = 2\pi \mathop \smallint \limits_c^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$
${\rm{UpperArea}}\left( T \right) = 2\pi \mathop \smallint \limits_{b - a}^{b + a} y\sqrt {1 + {{\left( {\dfrac{{ - \left( {y - b} \right)}}{{\sqrt {{a^2} - {{\left( {y - b} \right)}^2}} }}} \right)}^2}} {\rm{d}}y$
$ = 2\pi \mathop \smallint \limits_{b - a}^{b + a} y\sqrt {1 + \dfrac{{{{\left( {y - b} \right)}^2}}}{{{a^2} - {{\left( {y - b} \right)}^2}}}} {\rm{d}}y$
$ = 2\pi \mathop \smallint \limits_{b - a}^{b + a} y\sqrt {\dfrac{{{a^2}}}{{{a^2} - {{\left( {y - b} \right)}^2}}}} {\rm{d}}y$
$ = 2\pi \mathop \smallint \limits_{b - a}^{b + a} \dfrac{{ay}}{{\sqrt {{a^2} - {{\left( {b - y} \right)}^2}} }}{\rm{d}}y$
Since the total area of the torus is twice the area of the upper half, so
${\rm{Area}}\left( T \right) = 4\pi \mathop \smallint \limits_{b - a}^{b + a} \dfrac{{ay}}{{\sqrt {{a^2} - {{\left( {b - y} \right)}^2}} }}{\rm{d}}y$
(b) From part (a) we obtain
${\rm{Area}}\left( T \right) = 4\pi \mathop \smallint \limits_{b - a}^{b + a} \dfrac{{ay}}{{\sqrt {{a^2} - {{\left( {b - y} \right)}^2}} }}{\rm{d}}y$
Next, we compute this integral:
Write $u = b - y$. So, the integral becomes
${\rm{Area}}\left( T \right) = - 4\pi \mathop \smallint \limits_a^{ - a} \dfrac{{a\left( {b - u} \right)}}{{\sqrt {{a^2} - {u^2}} }}{\rm{d}}u$
${\rm{Area}}\left( T \right) = 4\pi \left[ {ab\mathop \smallint \limits_{ - a}^a \dfrac{1}{{\sqrt {{a^2} - {u^2}} }}{\rm{d}}u - a\mathop \smallint \limits_{ - a}^a \dfrac{u}{{\sqrt {{a^2} - {u^2}} }}{\rm{d}}u} \right]$
From the table of integral, we know that $\smallint \dfrac{1}{{\sqrt {{a^2} - {x^2}} }}{\rm{d}}x = {\sin ^{ - 1}}\dfrac{x}{a}$.
Write $t = {a^2} - {u^2}$. So, $dt = - 2udu$. The integral becomes
${\rm{Area}}\left( T \right) = 4\pi \left[ {ab\left( {{{\sin }^{ - 1}}\dfrac{u}{a}} \right)|_{ - a}^a + \dfrac{a}{2}\mathop \smallint \limits_0^0 \dfrac{1}{{\sqrt t }}{\rm{d}}t} \right]$
The second integral on the right-hand side vanishes, so
${\rm{Area}}\left( T \right) = 4\pi ab\left( {\dfrac{\pi }{2} + \dfrac{\pi }{2}} \right) = 4{\pi ^2}ab$
Hence, ${\rm{Area}}\left( T \right) = 4{\pi ^2}ab$.
