$\"\"$

\

The function is $\"\"$ .

\

Find the surface area of the region formed by revolving the region on the $\"\"$ .

\

Consider $\"\"$ .

\

Differentiate on each side with respect to $\"\"$ .

\

$\"\"$

\

Area of surface of revolution formed by revolving the graph of $\"\"$ about horizontal or vertical axis is

\

$\"\"$ .

\

Where $\"\"$ is continuous function on interval $\"\"$ and $\"\"$ is the distance between the

\

graph of $\"\"$ and axis of revolution.

\

Here $\"\"$ .

\

Substitute corresponding values.

\

Therefore, $\"\"$ .

\

$\"\"$ .

\

$\"\"$

\

Find the integral by substitution method.

\

Consider $\"\"$

\

Let $\"\"$

\

$\"\"$

\

Substitute corresponding values.

\

$\"\"$ .

\

Integral formula: $\"\"$ .

\

$\"\"$

\

Substitute back $\"\"$ .

\

$\"\"$

\

Substitute integral limits.

\

$\"\"$

\

$\"\"$ sq-units.

\

$\"\"$

\

Surface area of revolution is $\"\"$ sq-units.