QuestionCalculate the surface area from revolving $x=\frac{1}{3}(y^{2}+2)^{3/2}$ around the x-axis for $1 \leq y \leq 2$.

Studdy Solution
Assumptions1. The curve is given by $x=\frac{1}{3}\left(y^{}+\right)^{3 /}$ . The curve is revolved about the x-axis3. The range of y is from1 toThe formula for the surface area generated by revolving a curve $y=f(x)$ about the x-axis from $x=a$ to $x=b$ is given byA =\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^} dxHowever, in this case, we have $x$ as a function of $y$, so we need to modify the formula toA =\pi \int_c^d x \sqrt{1 + \left(\frac{dx}{dy}\right)^} dywhere $c$ and $d$ are the limits for $y$.
3First, we need to find the derivative of $x$ with respect to $y$, $\frac{dx}{dy}$.
$$\frac{dx}{dy} = \frac{d}{dy}\left(\frac{1}{3}\left(y^{}+\right)^{3 /}\right)$$4Use the chain rule to differentiate the function. The chain rule states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function.
\frac{dx}{dy} = \frac{1}{3} \cdot \frac{3}{}\left(y^{}+\right)^{1 /} \cdoty5implify the derivative.
$$\frac{dx}{dy} = y\left(y^{}+\right)^{1 /}$$6Now we need to find the square of the derivative and add1 to it.
1 + \left(\frac{dx}{dy}\right)^ =1 + \left(y\left(y^{}+\right)^{1 /}\right)^7implify the expression.
1 + \left(\frac{dx}{dy}\right)^ =1 + y^\left(y^{}+\right)8Now we need to find the square root of the above expression.
\sqrt{1 + \left(\frac{dx}{dy}\right)^} = \sqrt{1 + y^\left(y^{}+\right)}9Now we can substitute the values of $x$ and \sqrt{1 + \left(\frac{dx}{dy}\right)^} into the surface area formula.
A =\pi \int1^ \frac{1}{3}\left(y^{}+\right)^{3 /} \sqrt{1 + y^\left(y^{}+\right)} dy10Now, we need to evaluate this integral to find the surface area. This integral is quite complex and may require numerical methods or software to solve.