Question

Find the exact area of the surface obtained by rotating the curve about \({\rm{X}}\)- axis.

\(y = \) \(\frac{{{x^3}}}{6} + \frac{1}{{2x}},\;\;\;\frac{1}{2} \le 5x \le 1\)

Short answer

The exact area of the surface obtained by rotating the curve about \({\rm{X}}\)- axis is\( - \frac{{263\pi }}{{256}}\).

Step-by-step solution

Definition of area of the surface.

The space occupied by the object's surface is referred to as the surface area.

Given parameters.

The given parameters are: \(y = \) \(\frac{{{x^3}}}{6} + \frac{1}{{2x}},\;\;\;\frac{1}{2} \le 5x \le 1\)

Finding the exact surface area.

Consider the given equation and simplify,

To find the surface area of the curve when\(f(x)\)is rotated about the\(x\)-axis

Surface area,

\(\begin{aligned}{} &= \int_{1/2}^1 2 \pi y\sqrt {1 + {{\left( {{y^\prime }} \right)}^2}} dx\\ &= 2\pi \int_{1/2}^1 f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx{f^\prime }(x)\\ &= \underbrace {\frac{{6x\left( {4{x^3}} \right) - \left( {{x^4} + 3} \right)(6)}}{{{{(6x)}^2}}}}_{{\rm{Using Quotient Rule }}}\\ &= \frac{{x\left( {4{x^3}} \right) - \left( {{x^4} + 3} \right)}}{{6{x^2}}}\\ &= \frac{{4{x^4} - {x^4} - 3}}{{6{x^2}}} &= \frac{{{x^4} - 1}}{{2{x^2}}}\end{aligned}\)

Substitute \(f(x)\)and \({f^\prime }(x)\),

To get

\(\begin{aligned}{} &= 2\pi \int_{1/2}^1 {\left( {\frac{{{x^4} + 3}}{{6x}}} \right)} \sqrt {1 + {{\left( {\frac{{{x^4} - 1}}{{2{x^2}}}} \right)}^2}} dx\\ &= 2\pi \int_{1/2}^1 {\left( {\frac{{{x^4} + 3}}{{6x}}} \right)} \sqrt {1 + \frac{{{{\left( {{x^4} - 1} \right)}^2}}}{{4{x^4}}}} dx\\ &= 2\pi \int_{1/2}^1 {\left( {\frac{{{x^4} + 3}}{{6x}}} \right)} \sqrt {1 + \frac{{{x^8} - 2{x^4} + 1}}{{4{x^4}}}} dx\end{aligned}\)

Finding the exact surface area.

Evaluating further,

\(\begin{aligned}{} &- 2\pi \int_{1/2}^1 {\left( {\frac{{{x^4} + 3}}{{6x}}} \right)} \sqrt {\frac{{{x^8} + 4{x^4} - 2{x^4} + 1}}{{4{x^4}}}} dx\\ &- 2\pi \int_{1/2}^1 {\left( {\frac{{{x^4} + 3}}{{6x}}} \right)} \sqrt {\frac{{{x^8} + 2{x^4} + 1}}{{4{x^4}}}} dx\\ &- 2\pi \int_{1/2}^1 {\left( {\frac{{{x^4} + 3}}{{6x}}} \right)} \sqrt {\frac{{{{\left( {{x^4} + 1} \right)}^2}}}{{4{x^4}}}} dx\\ &- 2\pi \int_{1/2}^1 {\left( {\frac{{{x^4} + 3}}{{6x}}} \right)} \left( {\frac{{{x^4} + 1}}{{2{x^2}}}} \right)dx\\ &- 2\pi \int_{1/2}^1 {\frac{{{x^8} + 4{x^4} + 3}}{{12{x^3}}}} dx\end{aligned}\)

\(\begin{aligned}{} &- 2\pi \int_{1/2}^1 {\frac{{{x^8}}}{{12{x^3}}}} + \frac{{4{x^4}}}{{12{x^3}}} + \frac{3}{{12{x^3}}}dx\\ &- 2\pi \int_{1/2}^1 {\frac{{{x^5}}}{{12}}} + \frac{x}{3} + \frac{{{x^{ - 3}}}}{4}dx\\ &- 2\pi \left( {\frac{{{x^6}}}{{72}} + \frac{{{x^2}}}{6} - \frac{{{x^{ - 2}}}}{8}} \right)_{1/2}^1\\ &- 2\pi \left\{ {\left( {\frac{1}{{72}} + \frac{1}{6} - \frac{1}{8}} \right) - \left( {\frac{1}{{64 \times 72}} + \frac{1}{{24}} - \frac{4}{8}} \right)} \right\}\\ &- \frac{{263\pi }}{{256}}\end{aligned}\)

Hence, the required surface area is \( - \frac{{263\pi }}{{256}}\).