# Answer to Question #6471 in Calculus for marie

Question #6471

find the area of the surface generated by revolving the following curve about the x-axisy=sqrt(x^2+2), 0 is

Expert's answer

We'll use the following formula:

V = 2πr*[int](from 0 to sqrt(2))sqrt(x²+2)dx

Let's evaluate the integral:

[int]sqrt(x²+2)dx = 0.5*x*sqrt(x²+2)+asinh(0.5*sqrt(2)*x)

[int](from 0 to sqrt(2))sqrt(x²+2)dx = sqrt(2)+lg(1+sqrt(2))

r = [max](sqrt(x²+2)), where 0 <= x <= sqrt(2).

Obviously,

r = sqrt((sqrt(2))²+2) = sqrt(2+2) = 2.

So,

V = 2*π*2*(sqrt(2)+lg(1+sqrt(2))) = 4π(sqrt(2)+lg(1+sqrt(2))).

V = 2πr*[int](from 0 to sqrt(2))sqrt(x²+2)dx

Let's evaluate the integral:

[int]sqrt(x²+2)dx = 0.5*x*sqrt(x²+2)+asinh(0.5*sqrt(2)*x)

[int](from 0 to sqrt(2))sqrt(x²+2)dx = sqrt(2)+lg(1+sqrt(2))

r = [max](sqrt(x²+2)), where 0 <= x <= sqrt(2).

Obviously,

r = sqrt((sqrt(2))²+2) = sqrt(2+2) = 2.

So,

V = 2*π*2*(sqrt(2)+lg(1+sqrt(2))) = 4π(sqrt(2)+lg(1+sqrt(2))).

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