Question

(a) The ellipse

\(\frac{{{{\rm{x}}^{\rm{2}}}}}{{{{\rm{a}}^{\rm{2}}}}}{\rm{ + }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{{{\rm{b}}^{\rm{2}}}}}{\rm{ = 1}}\)

is rotated about the \({\rm{x - }}\)axis to form a surface called an ellipsoid, or prolate spheroid. Find the surface area of this ellipsoid.

(b) If the ellipse in part (a) is rotated about its minor axis (the \({\rm{y - }}\)axis), the resulting ellipsoid is called an oblate spheroid. Find the surface area of this ellipsoid.

Short answer

(a) The surface area is found to be \({\rm{S = 2\pi }}\left[ {\frac{{{{\rm{a}}^{\rm{2}}}{\rm{b}}}}{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{\rm{arcsin}}\left( {\frac{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{{\rm{a}}}} \right){\rm{ + }}{{\rm{b}}^{\rm{2}}}} \right]\).

(b)The surface area is found to be \({\rm{S = 2\pi }}{{\rm{a}}^{\rm{2}}}{\rm{ + }}\frac{{{\rm{2\pi a}}{{\rm{b}}^{\rm{2}}}}}{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{\rm{ln}}\left| {\frac{{{\rm{a + }}\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{{\rm{b}}}} \right|\).

Step-by-step solution

Concept Introduction

The area of an ellipse is specified by its major and minor axes, and the form of the ellipse is oval. The area of an ellipse\({\rm{ = }}\pi {\rm{ab}}\), where\({\rm{a}}\)and\({\rm{b}}\)are the lengths of the ellipse's semi-major and semi-minor axes, respectively. Ellipse is analogous to other portions of the conic section that are open in shape and unbounded, such as parabola and hyperbola.

Derivation of Equation

(a)

Recall from your elite conics knowledge, this is centred at the origin, has a horizontal major axis and\({\rm{a}}\)is the distancefrom the origin to a major vertex. So, double the result from integrating from\({\rm{0}}\)to\({\rm{a}}\).

For rotating about the\({\rm{x}}\)-axis, solve for\({\rm{y}}\)and keep the positive square root, since onlythe top half is needed. Then find\({\rm{y'}}\)for the integral later.

\(\frac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ + }}\frac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}{\text{ = 1, a > b}}\)

\(\frac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}{\text{ = 1 - }}\frac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}\)

\({{\text{y}}^{\text{2}}}{\text{ = }}{{\text{b}}^{\text{2}}}\left( {{\text{1 - }}\frac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}} \right){\text{ = }}\frac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} \right){\text{ factor out }}\frac{{\text{1}}}{{{{\text{a}}^{\text{2}}}}}\)

\({\text{y = }}\frac{{\text{b}}}{{\text{a}}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} {\text{ square root both sides}}\)

\({{\text{y}}^{\text{'}}}{\text{ = }}\frac{{\text{b}}}{{\text{a}}}{\text{x}}\frac{{\text{1}}}{{\text{2}}}{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} \right)^{{\text{ - 1/2}}}}{\text{( - 2x)}}\)

\({\text{ = - }}\frac{{{\text{bx}}}}{{{\text{a}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} }}\)

Substituting the values

Plug the stuff into the surface area formula (doubled) –

\({\text{S = 2}}\int_{\text{c}}^{\text{d}} {\text{2}} {\text{πy}}\sqrt {{\text{1 + }}{{\left( {{{\text{y}}^{\text{'}}}} \right)}^{\text{2}}}} {\text{dx}}\)

\({\text{ = 2}}\int_{\text{0}}^{\text{a}} {\text{2}} {\text{π}}\left( {\frac{{\text{b}}}{{\text{a}}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } \right)\sqrt {{\text{1 + }}{{\left( {{\text{ - }}\frac{{{\text{bx}}}}{{{\text{a}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} }}} \right)}^{\text{2}}}} {\text{dx}}\)

\({\text{ = }}\frac{{{\text{4πb}}}}{{\text{a}}}\int_{\text{0}}^{\text{a}} {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} } \sqrt {{\text{1 + }}\frac{{{{\text{b}}^{\text{2}}}{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} \right)}}} {\text{dx}}\)

\({\text{ = }}\frac{{{\text{4πb}}}}{{\text{a}}}\int_{\text{0}}^{\text{a}} {\sqrt {\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} \right)\left( {{\text{1 + }}\frac{{{{\text{b}}^{\text{2}}}{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}} \right)}}} \right)} } {\text{dx [}}\sqrt {\text{u}} \sqrt {\text{v}} {\text{ = }}\sqrt {{\text{uv}}} {\text{]}}\)

\({\text{ = }}\frac{{{\text{4πb}}}}{{\text{a}}}\int_{\text{0}}^{\text{a}} {\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{x}}^{\text{2}}}{\text{ + }}\frac{{{{\text{b}}^{\text{2}}}{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}} } {\text{dx}}\)

\({\text{ = }}\frac{{{\text{4πb}}}}{{\text{a}}}\int_{\text{0}}^{\text{a}} {\sqrt {\frac{{{{\text{a}}^{\text{4}}}{\text{ - }}{{\text{a}}^{\text{2}}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}} } {\text{dx common denominator}}\)

\({\text{ = }}\frac{{{\text{4πb}}}}{{\text{a}}}\int_{\text{0}}^{\text{a}} {\frac{{\text{1}}}{{\text{a}}}} \sqrt {{{\text{a}}^{\text{4}}}{\text{ - }}{{\text{a}}^{\text{2}}}{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}{{\text{x}}^{\text{2}}}} {\text{dx}}\)

\({\text{ = }}\frac{{{\text{4πb}}}}{{{{\text{a}}^2}}}\int_{\text{0}}^{\text{a}} {\sqrt {{{\text{a}}^{\text{4}}}{\text{ - (}}{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}{\text{)}}{{\text{x}}^{\text{2}}}} {\text{dx}}} \)

\({\text{ = }}\frac{{{\text{4πb}}}}{{{{\text{a}}^2}}}\int_{\text{0}}^{\text{a}} {\sqrt {{{\text{a}}^{\text{4}}}\left( {{\text{1 - }}\frac{{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)}}{{{{\text{a}}^{\text{4}}}}}{{\text{x}}^{\text{2}}}} \right)} {\text{dx factor out }}} {{\text{a}}^{\text{4}}}\)

\({\text{ = }}\frac{{{\text{4πb}}}}{{{{\text{a}}^2}}}\int_{\text{0}}^{\text{a}} {{{\text{a}}^2}\sqrt {{\text{1 - }}\frac{{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)}}{{{{\text{a}}^{\text{4}}}}}{{\text{x}}^{\text{2}}}} {\text{dx }}} \)

\({\text{ = 4πb}}\int_{\text{0}}^{\text{a}} {\sqrt {{\text{1 - }}\frac{{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)}}{{{{\text{a}}^{\text{4}}}}}{{\text{x}}^{\text{2}}}} {\text{dx }}} \)

Using Pythagoras Theorem

Carry out trigonometric substitution –

Let \(\frac{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{{{{\rm{a}}^{\rm{2}}}}}{\rm{x = sin\theta }} \to {\rm{x = }}\frac{{{{\rm{a}}^{\rm{2}}}}}{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{\rm{sin\theta }}\).

Then, it is\({\rm{dx = }}\frac{{{{\rm{a}}^{\rm{2}}}}}{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{\rm{cos\theta d\theta }}\).

Also, for later converting back to \({\rm{x}}\)–

\({\rm{\theta = arcsin}}\left( {\frac{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{{{{\rm{a}}^{\rm{2}}}}}{\rm{x}}} \right)\)

And forming a right triangle we have opposite =\({\rm{x}}\sqrt {\left( {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} \right)} \), hypotenuse =\({{\rm{a}}^2}\). So adjacent side is –

\({\text{adj = }}\sqrt {{{\left( {{{\text{a}}^{\text{2}}}} \right)}^{\text{2}}}{\text{ - }}{{\left( {{\text{x}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} } \right)}^{\text{2}}}} {\text{ = }}\sqrt {{{\text{a}}^{\text{4}}}{\text{ - }}{{\text{x}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} \)

\({\text{cos\θ= }}\frac{{\sqrt {{{\text{a}}^{\text{4}}}{\text{ - }}{{\text{x}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{a}}^{\text{2}}}}}\)

Now, on evaluating –

\({\text{4bπ }}\int_{\text{0}}^{\text{a}} {\sqrt {{\text{1 - }}\frac{{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)}}{{{{\text{a}}^{\text{4}}}}}{{\text{x}}^{\text{2}}}} } {\text{dx}}\)

\({\text{ = 4bπ}}\int_{\text{0}}^{{\text{x(a)}}} {\sqrt {{\text{1 - si}}{{\text{n}}^{\text{2}}}{\text{θ}}} } \frac{{{{\text{a}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{cosθdθ }}\)

\({\text{ = }}\frac{{{\text{4}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\int_{\text{0}}^{{\text{x(a)}}} {\sqrt {{\text{co}}{{\text{s}}^{\text{2}}}{\text{θ}}} } {\text{cosθdθ}}\)

\({\text{ = }}\frac{{{\text{4}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\int_{\text{0}}^{{\text{x(a)}}} {{\text{co}}{{\text{s}}^{\text{2}}}} {\text{θdθ}}\)

\({\text{ = }}\frac{{{\text{4}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\int_{\text{0}}^{{\text{x(a)}}} {\frac{{\text{1}}}{{\text{2}}}} {\text{(1 + cos2θ)dθ power reduction formula}}\)

\({\text{ = }}\frac{{{\text{2}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\int_{\text{0}}^{{\text{x(a)}}} {{\text{(1 + cos2θ)}}} {\text{dθ}}\)

\({\text{ = }}\frac{{{\text{2}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {{\text{θ+ }}\frac{{\text{1}}}{{\text{2}}}{\text{sin2θ}}} \right]_{\text{0}}^{{\text{x(a)}}}\)

\({\text{ = }}\frac{{{\text{2}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {{\text{θ+ }}\frac{{\text{1}}}{{\text{2}}}{\text{(2sinθcosθ)}}} \right]_{\text{0}}^{{\text{x(a)}}}{\text{ double angle formula}}\)

Substituting the values

Convert back to\({\rm{x}}\) using the stuff from the trig substitution box above –

\({\text{S = }}\frac{{{\text{2}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{[θ+ sinθcosθ]}}_{\text{0}}^{{\text{x(a)}}}\)

\({\text{ = }}\frac{{{\text{2}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {{\text{arcsin}}\left( {\frac{{{\text{x}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{a}}^{\text{2}}}}}} \right){\text{ + }}\frac{{{\text{x}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{a}}^{\text{2}}}}} \cdot \frac{{\sqrt {{{\text{a}}^{\text{4}}}{\text{ - }}{{\text{x}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{a}}^{\text{2}}}}}} \right]_{\text{0}}^{\text{a}}\)

\({\text{ = }}\frac{{{\text{2}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {{\text{arcsin}}\left( {\frac{{{\text{a}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{a}}^{\text{2}}}}}} \right){\text{ + }}\frac{{{\text{a}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{a}}^{\text{2}}}}} \cdot \frac{{\sqrt {{{\text{a}}^{\text{4}}}{\text{ - }}{{\text{a}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{a}}^{\text{2}}}}}{\text{ - (0 + 0)}}} \right]\)

\({\text{ = }}\frac{{{\text{2}}{{\text{a}}^{\text{2}}}{\text{bπ}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {{\text{arcsin}}\left( {\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{a}}}} \right){\text{ + }}\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \sqrt {{{\text{a}}^{\text{4}}}{\text{ - }}{{\text{a}}^{\text{4}}}{\text{ + }}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}} }}{{{{\text{a}}^{\text{3}}}}}} \right]\)

\({\text{ = 2π}}\left[ {\frac{{{{\text{a}}^{\text{2}}}{\text{b}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{arcsin}}\left( {\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{a}}}} \right){\text{ + }}\frac{{{{\text{a}}^{\text{2}}}{\text{b}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }} \cdot \frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \sqrt {{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}} }}{{{{\text{a}}^{\text{3}}}}}} \right]\)

\({\text{ = 2π}}\left[ {\frac{{{{\text{a}}^{\text{2}}}{\text{b}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{arcsin}}\left( {\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{a}}}} \right){\text{ + }}\frac{{{\text{b}} \cdot {\text{ab}}}}{{\text{a}}}} \right]\)

\({\text{ = 2π}}\left[ {\frac{{{{\text{a}}^{\text{2}}}{\text{b}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{arcsin}}\left( {\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{a}}}} \right){\text{ + }}{{\text{b}}^{\text{2}}}} \right]\)

Therefore, the value for surface area is \({\rm{S = 2\pi }}\left( {\frac{{{{\rm{a}}^{\rm{2}}}{\rm{b}}}}{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{\rm{arcsin}}\left( {\frac{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{{\rm{a}}}} \right){\rm{ + }}{{\rm{b}}^{\rm{2}}}} \right)\).

Derivation of Equation

(b)

Recall from your elite conics knowledge, this is centred at the origin, has a horizontal major axis and\({\rm{b}}\)is the distance from the origin to a minor vertex. So, double the result from integrating from\({\rm{0}}\)to\({\rm{b}}\).

For rotating about the\({\rm{y}}\)-axis, solve for\({\rm{x}}\)and keep the positive square root, since onlythe top half is needed. Then find\({\rm{x'}}\)for the integral later.

\(\frac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ + }}\frac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}{\text{ = 1, a > b about the y - axis}}\)

\(\frac{{{{\text{x}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ = 1 - }}\frac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}\)

\({{\text{x}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}\left( {{\text{1 - }}\frac{{{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}} \right){\text{ = }}\frac{{{{\text{b}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}\left( {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} \right){\text{ factor out }}\frac{{\text{1}}}{{{{\text{b}}^{\text{2}}}}}\)

\({\text{x = }}\frac{{\text{a}}}{{\text{b}}}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} {\text{ square root both sides}}\)

\({{\text{x}}^{\text{'}}}{\text{ = }}\frac{{\text{a}}}{{\text{b}}} \cdot \frac{{\text{1}}}{{\text{2}}}{\left( {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} \right)^{{\text{ - 1/2}}}}{\text{( - 2y)}}\)

\({\text{ = - }}\frac{{{\text{ay}}}}{{{\text{b}}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} }}\)

Substituting the values

Plug the stuff into the surface area formula (doubled) –

\({\text{S = 2}}\int_{\text{c}}^{\text{d}} {\text{2}} {\text{\pi x}}\sqrt {{\text{1 + }}{{\left( {{{\text{x}}^{\text{'}}}} \right)}^{\text{2}}}} {\text{dy}}\)

\({\text{ = 2}}\int_{\text{0}}^{\text{a}} {\text{2}} {\text{\pi }}\left( {\frac{{\text{a}}}{{\text{b}}}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} } \right)\sqrt {{\text{1 + }}{{\left( {{\text{ - }}\frac{{{\text{ay}}}}{{{\text{b}}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} }}} \right)}^{\text{2}}}} {\text{dy}}\)

Notice that this is like part (a) of this problem at the same point of the work, but with\({\rm{y}}\)instead of\({\rm{x}}\), and\({\rm{a}}\)and\({\rm{b}}\)swapped. However, we can't simply swap\({\rm{a}}\)and\({\rm{b}}\)because that results in\({\rm{a}}\)negativeinside the radical since\({\rm{a > b}}\).

\({\rm{S = 2\pi }}\left( {\frac{{{\rm{a}}{{\rm{b}}^{\rm{2}}}}}{{\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - }}{{\rm{a}}^{\rm{2}}}} }}{\rm{arcsin}}\left( {\frac{{\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - }}{{\rm{a}}^{\rm{2}}}} }}{{\rm{b}}}} \right){\rm{ + }}{{\rm{a}}^{\rm{2}}}} \right){\rm{ (wrong!)}}\)

\({\rm{S = }}\frac{{{\rm{4\pi a}}}}{{\rm{b}}}\int_{\rm{0}}^{\rm{b}} {\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} } \sqrt {{\rm{1 + }}\frac{{{{\rm{a}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}}}{{{{\rm{b}}^{\rm{2}}}\left( {{{\rm{b}}^{\rm{2}}}{\rm{ - }}{{\rm{y}}^{\rm{2}}}} \right)}}} {\rm{dy}}\)

\({\text{ = }}\frac{{{\text{4πa}}}}{{\text{b}}}\int_{\text{0}}^{\text{b}} {\sqrt {\left( {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} \right)\left( {{\text{1 + }}\frac{{{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}\left( {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}} \right)}}} \right)} } {\text{dy [}}\sqrt {\text{u}} \sqrt {\text{v}} {\text{ = }}\sqrt {{\text{uv}}} {\text{]}}\)

\({\text{ = }}\frac{{{\text{4πa}}}}{{\text{b}}}\int_{\text{0}}^{\text{b}} {\sqrt {{{\text{b}}^{\text{2}}}{\text{ - }}{{\text{y}}^{\text{2}}}{\text{ + }}\frac{{{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}} } {\text{dy}}\)

\({\text{ = }}\frac{{{\text{4πa}}}}{{\text{b}}}\int_{\text{0}}^{\text{b}} {\frac{{\text{1}}}{{{{\text{b}}^{\text{2}}}}}} \sqrt {{{\text{b}}^{\text{4}}}{\text{ - }}{{\text{b}}^{\text{2}}}{{\text{y}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}} {\text{dy}}\)

\({\text{ = }}\frac{{{\text{4πa}}}}{{{{\text{b}}^{\text{2}}}}}\int_{\text{0}}^{\text{b}} {\sqrt {{{\text{b}}^{\text{4}}}{\text{ + (}}{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}{\text{)}}{{\text{y}}^{\text{2}}}} {\text{dy}}} \)

\({\text{ = }}\frac{{{\text{4πa}}}}{{{{\text{b}}^{\text{2}}}}}\int_{\text{0}}^{\text{b}} {\sqrt {{{\text{b}}^{\text{4}}}\left( {{\text{1 + }}\frac{{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)}}{{{{\text{b}}^{\text{4}}}}}{{\text{y}}^{\text{2}}}} \right)} {\text{dy factor out }}} {{\text{b}}^{\text{4}}}\)

\({\text{ = }}\frac{{{\text{4πa}}}}{{{{\text{b}}^{\text{2}}}}}\int_{\text{0}}^{\text{b}} {{{\text{b}}^{\text{2}}}\sqrt {{\text{1 + }}\frac{{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)}}{{{{\text{b}}^{\text{4}}}}}{{\text{y}}^{\text{2}}}} {\text{dy }}} \)

\({\text{ = 4πa}}\int_{\text{0}}^{\text{b}} {\sqrt {{\text{1 + }}\frac{{\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)}}{{{{\text{b}}^{\text{4}}}}}{{\text{y}}^{\text{2}}}} {\text{dy }}} \)

Using Pythagoras Theorem

Carry out trigonometric substitution –

\({\text{y = }}\frac{{{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{tan t}}\)

\({\text{dy = }}\frac{{{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{se}}{{\text{c}}^{\text{2}}}{\text{t dt}}\)

Also, for later converting back to \({\rm{y}}\)–

\({\text{tant = }}\frac{{{\text{y}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}{\text{ = }}\frac{{{\text{opp}}}}{{{\text{adj}}}}\)

\({\text{hyp = }}\sqrt {{\text{ad}}{{\text{j}}^{\text{2}}}{\text{ + op}}{{\text{p}}^{\text{2}}}} {\text{ = }}\sqrt {{{\left( {{{\text{b}}^{\text{2}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {{\text{y}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} } \right)}^{\text{2}}}} \)

\({\text{ = }}\sqrt {{{\text{b}}^{\text{4}}}{\text{ + }}{{\text{y}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} \)

\({\text{sect = }}\frac{{{\text{hyp}}}}{{{\text{adj}}}}{\text{ = }}\frac{{\sqrt {{{\text{b}}^{\text{4}}}{\text{ + }}{{\text{y}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}\)

Change back to the original variable later, so leave the\({\rm{t}}\)limits blank.

\({\text{S = 4πa}}\int {\sqrt {{\text{1 + ta}}{{\text{n}}^{\text{2}}}{\text{t}}} } \frac{{{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{se}}{{\text{c}}^{\text{2}}}{\text{t dt}}\)

\({\text{ = }}\frac{{{\text{4πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\int {\sqrt {{\text{se}}{{\text{c}}^{\text{2}}}{\text{t}}} } {\text{ se}}{{\text{c}}^{\text{2}}}{\text{t dt}}\)

\({\text{ = }}\frac{{{\text{4πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\int {{\text{se}}{{\text{c}}^{\text{3}}}} {\text{t dt}}\)

For the integration of \({\rm{sec t}}\),

\({\text{ = }}\frac{{{\text{4πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {\frac{{\text{1}}}{{\text{2}}}{\text{(sec t tan t + ln|sec t + tan t|)}}} \right]\)

\({\text{ = }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{[sec t tan t + ln|sec t + tan t|]}}\)

Substituting the values

Convert back to \({\rm{y}}\) using the stuff from the trig substitution box above –

\({\text{ = }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {\frac{{\sqrt {{{\text{b}}^{\text{4}}}{\text{ + }}{{\text{y}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}\frac{{{\text{y}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + ln}}\left| {\frac{{\sqrt {{{\text{b}}^{\text{4}}}{\text{ + }}{{\text{y}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + }}\frac{{{\text{y}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}} \right|} \right]_{\text{0}}^{\text{b}}\)

\({\text{ = }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {\frac{{\sqrt {{{\text{b}}^{\text{4}}}{\text{ + (b}}{{\text{)}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}\frac{{{\text{(b)}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + ln}}\left| {\frac{{\sqrt {{{\text{b}}^{\text{4}}}{\text{ + (b}}{{\text{)}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + }}\frac{{{\text{(b)}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}} \right|} \right.\)

\(\left. {{\text{ - }}\left( {\frac{{\sqrt {{{\text{b}}^{\text{4}}}{\text{ + (0}}{{\text{)}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}\frac{{{\text{(0)}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + ln}}\left| {\frac{{\sqrt {{{\text{b}}^{\text{4}}}{\text{ + (0}}{{\text{)}}^{\text{2}}}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + }}\frac{{{\text{(0)}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}} \right|} \right)} \right]\)

\({\text{ = }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {\frac{{{\text{b}}\sqrt {{{\text{b}}^{\text{2}}}{\text{ + }}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{b}}}{\text{ + ln}}\left| {\frac{{{\text{b}}\sqrt {{{\text{b}}^{\text{2}}}{\text{ + }}\left( {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} \right)} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + }}\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{b}}}} \right|{\text{ - (0 + 0)}}} \right]\)

\({\text{ = }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {\frac{{\sqrt {{{\text{a}}^{\text{2}}}} }}{{\text{b}}}\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{b}}}{\text{ + ln}}\left| {\frac{{\sqrt {{{\text{a}}^{\text{2}}}} }}{{\text{b}}}{\text{ + }}\frac{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{b}}}} \right|} \right]\)

\({\text{ = }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\left[ {\frac{{{\text{a}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + ln}}\left| {\frac{{{\text{a + }}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{b}}}} \right|} \right]\)

\({\text{ = }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}\frac{{{\text{a}}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{{{\text{b}}^{\text{2}}}}}{\text{ + }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{ln}}\left| {\frac{{{\text{a + }}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{b}}}} \right|\)

\({\text{ = 2π}}{{\text{a}}^{\text{2}}}{\text{ + }}\frac{{{\text{2πa}}{{\text{b}}^{\text{2}}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{\text{ln}}\left| {\frac{{{\text{a + }}\sqrt {{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}} }}{{\text{b}}}} \right|\)

Therefore, the value for surface area is \({\rm{S = 2\pi }}{{\rm{a}}^{\rm{2}}}{\rm{ + }}\frac{{{\rm{2\pi a}}{{\rm{b}}^{\rm{2}}}}}{{\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{\rm{ln}}\left| {\frac{{{\rm{a + }}\sqrt {{{\rm{a}}^{\rm{2}}}{\rm{ - }}{{\rm{b}}^{\rm{2}}}} }}{{\rm{b}}}} \right|\).