Question

The given curve is rotated about the \({\rm{x}}\)-axis. Find the area of the resulting surface.

\(x = \sqrt {{a^2} - {y^2}} ,\;\;\;0 \le y \le a/2\)

Short answer

The area of the resulting surface is\(\pi {a^2}\).

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: \(x = \sqrt {{a^2} - {y^2}} ,\;\;\;0 \le y \le a/2\)

Finding the area of resulting surface

Let us consider the given parameters,

Surface Area after rotating about \(y\)-axis is:

\(x = \sqrt {{a^2} - {y^2}} \)

We define the surface area of the surface acquired by rotating the curve, where\(f\)is positive and has a continuous derivative.

\(y = (x),a \le x \le b\), about the\(y\)-axis as:

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

We are differentiating (1) with respect to\(x\).

\(\begin{aligned} &= 2\pi \int_0^{a/2} {\sqrt {{a^2} - {y^2}} } \sqrt {\frac{{{a^2} - {y^2} + {y^2}}}{{{a^2} - {y^2}}}} dy\\ &= 2\pi \int_0^{a/2} {\sqrt {{a^2} - {y^2}} } \sqrt {\frac{{{a^2}}}{{{a^2} - {y^2}}}} dy\end{aligned}\)

\(\begin{aligned} &= 2\pi \int_0^{a/2} {\sqrt {\frac{{\left( {{a^2} - {y^2}} \right){a^2}}}{{{a^2} - {y^2}}}} } dy\\ &= 2\pi \int_0^{a/2} {\sqrt {{a^2}} } dy\end{aligned}\)

\(\begin{aligned} &= 2\pi \int_0^{a/2} a dy\\ &= 2\pi (ay)_0^{a/2}\end{aligned}\)

\(\begin{aligned} &= 2\pi \cdot \frac{{{a^2}}}{2}\\ &= \pi {a^2}\end{aligned}\)

Therefore, the area of the surface is\(\pi {a^2}\).