Find the area of the surface obtained by rotating the curve y= \sqrt(3 x) from x=0 to x = 4 about the x-axis. Answer in square units.

Hello!

There is the formula for the area of a surface of revolution: if a function f(x)  is considered for x  from a  to b  and its graph is rotated over the x-axis, then the surface area is

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

In our question a = 0 ,  b = 4 ,  f ( x ) = sqrt ( 3 x ) ,  f ' ( x ) = sqrt ( 3 ) / ( 2sqrt ( x ) ) .

This way, the area is

A = int_0^4 ( 2 pi sqrt ( 3 x ) sqrt ( 1 + 3 / (4x) ) ) dx ​= int_0^4 ( pi sqrt(3) sqrt ( 4 x + 3 ) ) dx .

To make it even simpler, substitute u = 4 x + 3 ,  du = 4 dx ,  u  from 3  to 19 .  Then we obtain

A = int_3^19 ( pi sqrt(3)/4 sqrt ( u ) ) du = pi sqrt(3)/4 ( 2 / 3 ) ( u^( 3 / 2 ) )_( u = 3 )^19 =

= pi / (2sqrt(3)) ( 19^( 3 / 2 ) - 3^(3/2) ) = pi/6 (19 sqrt(57) - 9).

It is approximately 70.4 square units.

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