Question

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

\(y = 1 - {x^2},\;\;\;0 \le x \le 1\)

Short answer

The area of the resulting surface is\(\frac{\pi }{6}(5\sqrt 5 - 1)\).

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: \(y = 1 - {x^2},\;\;\;0 \le x \le 1\)

Finding the area of resulting surface.

Let us consider the given parameters,

\(y = 1 - {x^2}\)………….(1)

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

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

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

\(\begin{aligned}\frac{{dy}}{{dx}} &= - 2x\\{\left( {\frac{{dx}}{{dy}}} \right)^2} &= 4{x^2}\\s &= \int_0^1 2 \pi x\sqrt {1 + 4{x^2}} dxd\end{aligned}\)

Let\(u = 1 + 4{x^2} \to du = 8xdx\)

Changing the limits of integration

\(\begin{aligned}a = 5\;\;\;,b = 1s\\ = \frac{\pi }{4}\int_1^5 {\sqrt u } du\\ = \frac{\pi }{4}\left( {\frac{2}{3}{u^{3/2}}} \right)_1^5\\ = \frac{\pi }{6}(5\sqrt 5 - 1)\end{aligned}\)

Therefore, the area of the surface is \(\frac{\pi }{6}(5\sqrt 5 - 1)\).