what is the area of surface of revelution of function x= 1/3(y^2+2)^3/2 between the 1 <= y <= 2 ?

To find the surface area of revolution for the given function, we can use the formula for the surface area of revolution:

\(A = \int_{a}^{b}2\pi f(y)\sqrt{1+[f'(y)]^2}dy\)

Here, \(f(y) = \frac{1}{3}(y^2 + 2)^{\frac{3}{2}}\) represents the function we are revolving, and \(a = 1\) and \(b = 2\) represent the limits of integration.

First, let's find the derivative of the function \(f(y)\) with respect to \(y\).

\(f'(y) = \frac{d}{dy}\left(\frac{1}{3}(y^2 + 2)^{\frac{3}{2}}\right)\)

To find the derivative, we can use the chain rule. Let's define \(u = y^2 + 2\) so that \(f(y) = \frac{1}{3}u^{\frac{3}{2}}\).

Applying the chain rule, we have:

\(f'(y) = \frac{1}{3}\cdot\frac{d}{dy}\left(u^{\frac{3}{2}}\right)\)

Using the power rule, we find:

\(f'(y) = \frac{1}{3}\cdot\frac{3}{2} u^{\frac{1}{2}}\cdot\frac{du}{dy}\)

Substituting \(u = y^2 + 2\) and \(\frac{du}{dy} = 2y\), we get:

\(f'(y) = \frac{1}{2}(y^2 + 2)^{\frac{1}{2}}\cdot 2y\)

Simplifying, we have:

\(f'(y) = y(y^2 + 2)^{\frac{1}{2}}\)

Now we can substitute the function \(f(y)\) and its derivative \(f'(y)\) into the surface area formula:

\(A = \int_{1}^{2}2\pi\left(\frac{1}{3}(y^2 + 2)^{\frac{3}{2}}\right)\sqrt{1+\left[y(y^2 + 2)^{\frac{1}{2}}\right]^2}dy\)

Simplifying further:

\(A = \frac{2\pi}{3}\int_{1}^{2}(y^2 + 2)^{\frac{3}{2}}\sqrt{1+y^2(y^2 + 2)}dy\)

To solve this integral, you can use various techniques such as substitution, integration by parts, or trigonometric identities. Once you evaluate the definite integral, you will obtain the surface area of revolution for the given function between \(y = 1\) and \(y = 2\).