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Expert-verified Found in: Page 807 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # 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.

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.

See the step by step solution

## Step 1: 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.

## Step 2: 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. ### Want to see more solutions like these? 