## Related questions with answers

Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the y-axis. $y=\frac{x^{2}}{2}+4$, $0 \leq x \leq 2$

Solution

Verified$\color{#4257b2}{y=\frac{x^{2}}{2}+4, \quad 0 \leq x \leq 2}$

Differentiate with respect to $x$:

$\begin{align*} f^{\prime}(x) &=\frac{d}{d x} [\frac{x^{2}}{2}+4] \\ &=\frac{2 x}{2} \\ &=x \end{align*}$

Then the area of surface of revolution is:

$\begin{align*} S &=2 \pi \int_{0}^{2}(x)\left(\sqrt{1+(x)^{2}} \ d x\right) \\ &=2 \pi \int_{0}^{2}(x) \sqrt{1+x^{2}} \ d x \\ &=2 \pi\left[\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3}\right]_{0}^{2} \\ &=\frac{2 \pi}{3}\left(5^{\frac{3}{2}}-1\right) \\ &=21.3 \end{align*}$

$\color{#c34632}{S=21.3}$

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