There is formula for computing area of a surface of revolution. This formula is obtained by approximating the function (and therefore the surface) by small parts and adding the approximate results. So, it is not a surprise that some integral arises:

`A = int_a^b ( 2 pi f ( x...

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There is formula for computing area of a surface of revolution. This formula is obtained by approximating the function (and therefore the surface) by small parts and adding the approximate results. So, it is not a surprise that some integral arises:

`A = int_a^b ( 2 pi f ( x ) sqrt ( 1 + ( f ' ( x ) )^2 ) ) dx .`

In our task `a = 0 , ` `b = pi , ` `f ( x ) = 3 / 2 sin x . ` Find the derivative: `f ' ( x ) = 3 / 2 cos x . ` This way, we need to compute the integral

`A = int_0^pi ( 3 pi sin x sqrt ( 1 + 9 / 4 cos^2 x ) ) dx .`

It can be simplified by the substitution `u = 3 / 2 cos x , ` then `u ` is changing from `3 / 2 ` to `- 3 / 2 , ` `du = - 3 / 2 sin x dx . ` We obtain

`A = - int_( 3 / 2 )^( -3 / 2 ) (2 pi sqrt(1+u^2)) du = 2 pi int_( -3 / 2 )^( 3 / 2 ) sqrt(1+u^2) du .`

The last integral is more or less a table one, `int sqrt(1+u^2) du = 1/2(u sqrt(1+u^2)+ln(u+sqrt(1+u^2))). ` Substitute to the definite integral we have and use the fact that the function under the integral is even:

`A = 2pi (3/2 sqrt(1+9/4)+ln(3/2+sqrt(1+9/4))).`

It is approximately **24.5** square units.