Question

Find the exact area of the surface obtained by rotating the curve about \(x\)-axis.

\({\rm{x}} = \frac{1}{3}{\left( {{y^2} + 2} \right)^{3/2}},\;\;\;1 \le {\rm{y}} \le 2\)

Short answer

The exact area of the surface obtained by rotating the curve about \({\rm{X}}\)- axis is \( - \frac{{21\pi }}{2}\).

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: \(x = \frac{1}{3}{\left( {{y^2} + 2} \right)^{3/2}},\;\;\;1 \le y \le 2\)

Finding the exact surface area.

Consider the given parameters and simplify,

\(x - \frac{1}{3}{\left( {{y^2} + 2} \right)^{3/2}}\)….……… (1)

Where \(f\)is positive and has a continuous derivative, we define the surface area of the surface obtained by rotating the curve\(y - (x),a \le x \le b\), about the \(x\)-axis as:

\(s - \int_a^b 2 \pi y\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy\)

Differentiating (1) with respect to\(x\),

\(\begin{aligned}{}&\frac{{dx}}{{dy}} - \frac{1}{2}{\left( {{y^2} + 2} \right)^{1/2}}.2y\\&\frac{{dx}}{{dy}} - y{\left( {{y^2} + 2} \right)^{1/2}}\\&\frac{{dx}}{{dy}} - {y^2}\left( {{y^2} + 2} \right) - {y^4} + 2{y^2}\end{aligned}\)

\(\begin{aligned}{} &- \int_1^2 2 \pi y\sqrt {1 + {y^4} + 2{y^2}} dy\\ &- 2\pi \int_1^2 y \sqrt {{{\left( {{y^2} + 1} \right)}^2}} dy\\ &- 2\pi \int_1^2 y \left( {{y^2} + 1} \right)dy\\ &- 2\pi \int_1^2 {\left( {{y^3} + y} \right)} dy\\ &- 2\pi \left( {\frac{1}{4}{y^4} + \frac{1}{2}{y^2}} \right)_1^2\\ &- 2\pi \left( {4 + 2 - \frac{1}{4} - \frac{1}{2}} \right)\\ &- \frac{{21\pi }}{2}\end{aligned}\)
Therefore, the surface of the area =\( - \frac{{21\pi }}{2}\).