Question

To verify: The formula \(A = 2\pi \int_a^b f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\) by using in the definition 6 and the parametric equations for a surface of revolution.

Short answer

The formula \(10A = 2\pi \int_a^b f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\) is verified by using the definition 6 and the parametric equations for a surface of revolution.

Step-by-step solution

Use parametric equations for surface of revolution

It is known that, the parametric equations for surface of revolution is \(x = x,y = f(x)\cos \theta \) and \(z = f(x)\sin \theta \) Also, from the definition 6 it is known that, .

Use the parametric equations for surface of revolution and the definition 6 and verify the formula 10 \(A = 2\pi \int_a^b f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\) as follows.

Consider the surface which is obtained by rotating the curve \(y = f(x)\), where \(a \le x \le b\) about the \(x\)-axis. In order to define the surface area, divide the interval (\(a,b\)) into \(n\) subintervals such as the endpoints are \({x_0},{x_1},{x_2}, \ldots ,{x_n}\).

That implies, the surface area is \(2\pi \frac{{{y_{i - 1}} + {y_i}}}{2}\left| {{P_{i - 1}} - P} \right|\) as \(y = f(x)\) rotated about \(x\)-axis, where \(f\left( {x_i^*} \right) = \frac{{{y_{i - 1}} + {y_i}}}{2}\left| {{P_{i - 1}} - P} \right|\).

Let \(2\pi \frac{{{y_{i - 1}} + {y_i}}}{2}\left| {{P_{i - 1}} - P} \right|\) be equation (1).

Here, \(\left| {{P_{i - 1}} - P} \right| = \sqrt {1 + {{\left( {{f^\prime }\left( {x_i^*} \right)} \right)}^2}} \Delta x\), where \(x_i^*\) is some number between \(\left( {{x_{i - 1}},{x_i}} \right)\).

Note that, the value of \(\Delta x\) is small.

As the function \(f\) is continuous,

\(\begin{array}{c}{y_i} = f\left( {{x_i}} \right) \approx f\left( {x_i^*} \right){y_{i - 1}}\\ = f\left( {{x_{i - 1}}} \right) \approx f\left( {{x_i}} \right)\end{array}\)

Now equation\((1)\)implies,

\(2\pi \frac{{{y_{i - 1}} + {y_i}}}{2}\left| {{P_{i - 1}} - P} \right| \approx 2\pi f\left( {x_i^*} \right)\sqrt {1 + {{\left( {{f^\prime }\left( {x_i^*} \right)} \right)}^2}} \Delta x.\)

Ignore \(\Delta x\) as it is too small.

Since the value of \(n \to \infty ,g(x) = 2\pi f(x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} \).

That implies,

\(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n 2 \pi f(x)\sqrt {1 + {{\left( {{f^\prime }\left( {x_i^*} \right)} \right)}^2}} \Delta x = \int_a^b 2 \pi f(x)\sqrt {1 + {{\left( {{f^\prime }\left( {{x_i}} \right)} \right)}^2}} .\)

Note that, \(f\) is positive and it has a continuous derivative.

So, the surface area obtained by rotating the curve \(y = f(x)\), where \(a \le x \le b\) about the \(x\)-axis is \(A = 2\pi \int_a^b f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\)

Hence, the formula \(10A = 2\pi \int_a^b f (x)\sqrt {1 + {{\left( {{f^\prime }(x)} \right)}^2}} dx\) is verified by using the definition 6 and the parametric equations for a surface of revolution.