find the exact area of the surface obtained by rotating the curve about the x - axis.\ny =…

Explanation:

Step 1: Find the derivative of ( y )

Given ( y=\frac{x^{3}}{4}+\frac{1}{3x}=\frac{x^{3}}{4}+\frac{1}{3}x^{-1} ). Using the power rule ( (x^{n})^\prime = nx^{n - 1} ), we have ( y^\prime=\frac{3x^{2}}{4}-\frac{1}{3}x^{-2}=\frac{3x^{2}}{4}-\frac{1}{3x^{2}} ).

Step 2: Calculate ( 1+(y^\prime)^{2} )

[ \begin{align*} 1+(y^\prime)^{2}&=1+\left(\frac{3x^{2}}{4}-\frac{1}{3x^{2}}\right)^{2}\ &=1+\frac{9x^{4}}{16}-2\times\frac{3x^{2}}{4}\times\frac{1}{3x^{2}}+\frac{1}{9x^{4}}\ &=1+\frac{9x^{4}}{16}-\frac{1}{2}+\frac{1}{9x^{4}}\ &=\frac{9x^{4}}{16}+\frac{1}{2}+\frac{1}{9x^{4}}\ &=\left(\frac{3x^{2}}{4}+\frac{1}{3x^{2}}\right)^{2} \end{align*} ]

Step 3: Use the surface - area formula

The formula for the surface area ( S ) of a curve ( y = f(x) ) rotated about the ( x ) - axis is ( S=\int_{a}^{b}2\pi y\sqrt{1+(y^\prime)^{2}}dx ), where ( a=\frac{1}{2} ), ( b = 1 ), ( y=\frac{x^{3}}{4}+\frac{1}{3x} ), and ( \sqrt{1+(y^\prime)^{2}}=\frac{3x^{2}}{4}+\frac{1}{3x^{2}} ). [ \begin{align*} S&=\int_{\frac{1}{2}}^{1}2\pi\left(\frac{x^{3}}{4}+\frac{1}{3x}\right)\left(\frac{3x^{2}}{4}+\frac{1}{3x^{2}}\right)dx\ &=2\pi\int_{\frac{1}{2}}^{1}\left(\frac{3x^{5}}{16}+\frac{x^{3}}{12x^{2}}+\frac{3x^{2}}{12x}+\frac{1}{9x^{3}}\right)dx\ &=2\pi\int_{\frac{1}{2}}^{1}\left(\frac{3x^{5}}{16}+\frac{x}{12}+\frac{1}{4x}+\frac{1}{9x^{3}}\right)dx\ &=2\pi\left[\frac{3x^{6}}{96}+\frac{x^{2}}{24}+\frac{1}{4}\ln x-\frac{1}{18x^{2}}\right]_{\frac{1}{2}}^{1}\ \end{align*} ] Evaluate the definite integral: [ \begin{align*} &2\pi\left[\left(\frac{3(1)^{6}}{96}+\frac{(1)^{2}}{24}+\frac{1}{4}\ln(1)-\frac{1}{18(1)^{2}}\right)-\left(\frac{3(\frac{1}{2})^{6}}{96}+\frac{(\frac{1}{2})^{2}}{24}+\frac{1}{4}\ln(\frac{1}{2})-\frac{1}{18(\frac{1}{2})^{2}}\right)\right]\ &=2\pi\left[\left(\frac{3}{96}+\frac{1}{24}+0 - \frac{1}{18}\right)-\left(\frac{3}{96\times64}+\frac{1}{96}-\frac{1}{4}\ln2-\frac{4}{18}\right)\right]\ &=2\pi\left[\left(\frac{3 + 4}{96}-\frac{1}{18}\right)-\left(\frac{3}{6144}+\frac{1}{96}-\frac{1}{4}\ln2-\frac{4}{18}\right)\right]\ &=2\pi\left[\frac{7}{96}-\frac{1}{18}-\frac{3}{6144}-\frac{1}{96}+\frac{1}{4}\ln2+\frac{4}{18}\right]\ &=2\pi\left[\frac{7 - 1}{96}+\frac{4 - 1}{18}-\frac{3}{6144}+\frac{1}{4}\ln2\right]\ &=2\pi\left[\frac{6}{96}+\frac{3}{18}-\frac{3}{6144}+\frac{1}{4}\ln2\right]\ &=2\pi\left[\frac{1}{16}+\frac{1}{6}-\frac{3}{6144}+\frac{1}{4}\ln2\right]\ &=2\pi\left[\frac{384 + 1024- 3}{6144}+\frac{1}{4}\ln2\right]\ &=2\pi\left[\frac{1405}{6144}+\frac{1}{4}\ln2\right]\ &=\frac{1405\pi}{3072}+\frac{\pi}{2}\ln2 \end{align*} ]

Answer:

(\frac{1405\pi}{3072}+\frac{\pi}{2}\ln2)