Find the exact area of the surface obtained by rotating the given curve about the x-axis. y=sqrt{x^2 + 1} , 0 is less than or equal to x is less...

To determine the area of the solid of revolution
determined by rotating the curve around x axis, we'll use the
formula:

A = 2`pi` `int`
f(x)*sqrt{1+[f'(x)]^2}dx

In the given case, the function is
y = f(x) = sqrt(x^2 + 1) => f'(x) = 2x/2sqrt(x^2 + 1) = x/sqrt(x^2 + 1) and the
limits of integration are 0 to 3.

A = 2`pi` `int` sqrt(x^2
+ 1)*sqrt[1 + x^2/(x^2 + 1)]dx

A = 2`pi` `int` sqrt(2x^2 +
1)dx

We'll calculate the integral of irrational function
using the formula:

`int` r dx = [a^2*ln(x+r) + x*r]/2,
where r = sqrt(x^2 + a^2)

sqrt(2x^2 + 1) = sqrt2*sqrt(x^2 +
1/2)

Considering a = 1/sqrt2, we'll
get:

`int` sqrt(2x^2 + 1)dx = sqrt2*{ln[x+sqrt(x^2 +
1/2)]/2 + x*sqrt(x^2 + 1/2)}/2 + C

To calculate the
definite integral, we'll use Leibniz Newton formula:

`int`
sqrt(2x^2 + 1)dx = F(3) - F(0)

F(3) = sqrt2*{ln[3+sqrt(3^2
+ 1/2)]/2 + 3*sqrt(3^2 + 1/2)}/2

F(3) =
(sqrt2/2)*{ln[3+sqrt(19/2)]/2 + 3sqrt(19/2)}

F(0) =
sqrt2*{ln[0+sqrt(0^2 + 1/2)]/2 + 0*sqrt(0^2 + 1/2)}/2

F(0)
= (sqrt2/2)*{ln[sqrt(1/2)]/2}

F(3) - F(0) =
(sqrt2/2)*{ln[3+sqrt(19/2)]/2 + 3sqrt(19/2) -
ln[sqrt(1/2)]/2}

We'll use the quotient formula or
logarithms:

ln a - ln b = ln
(a/b)

F(3) - F(0) =
(sqrt2/2)*{ln[(3+sqrt(19/2))]/sqrt(1/2)] +
3sqrt(19/2)}

The area of the surface of the
solid of revolution determined by the given rotating the curve around x axis is F(3) -
F(0) = (sqrt2/2)*{ln[(3+sqrt(19/2))]/sqrt(1/2)] +
3sqrt(19/2)}.