an oil storage tank can be described as the volume generated by revolving the area bounded by y = 3/(16 +…

Explanation:

Step1: Recall volume - of - revolution formula

The formula for the volume $V$ of the solid generated by revolving the curve $y = f(x)$ about the $x$-axis from $x=a$ to $x = b$ is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $f(x)=\frac{3}{(16 + x^{2})^{\frac{1}{4}}}$, $a = 0$, and $b = 5$.

Step2: Set up the integral

$V=\pi\int_{0}^{5}\left(\frac{3}{(16 + x^{2})^{\frac{1}{4}}}\right)^{2}dx=\pi\int_{0}^{5}\frac{9}{(16 + x^{2})^{\frac{1}{2}}}dx$.

Step3: Use the integral formula $\int\frac{1}{\sqrt{a^{2}+x^{2}}}dx=\ln(x+\sqrt{x^{2}+a^{2}})+C$

We know that $\int\frac{9}{\sqrt{16 + x^{2}}}dx=9\int\frac{1}{\sqrt{4^{2}+x^{2}}}dx$. Since $\int\frac{1}{\sqrt{a^{2}+x^{2}}}dx=\ln(x+\sqrt{x^{2}+a^{2}})+C$ with $a = 4$, we have $9\int\frac{1}{\sqrt{4^{2}+x^{2}}}dx=9\ln(x+\sqrt{x^{2}+16})+C$.

Step4: Evaluate the definite - integral

$V=\pi\left[9\ln(x+\sqrt{x^{2}+16})\right]_{0}^{5}=\pi\left(9\ln(5+\sqrt{25 + 16})-9\ln(0+\sqrt{0 + 16})\right)=\pi\left(9\ln(5+\sqrt{41})-9\ln(4)\right)$. Using the property of logarithms $\ln m-\ln n=\ln\frac{m}{n}$, we get $V = 9\pi\ln\frac{5+\sqrt{41}}{4}$.

Step5: Calculate the numerical value

$V=9\pi\ln\frac{5+\sqrt{41}}{4}\approx9\pi\ln\frac{5 + 6.4031}{4}=9\pi\ln\frac{11.4031}{4}=9\pi\ln(2.8508)$. $V\approx9\times3.1416\times1.0446\approx29.6799$.

Answer:

$29.6799$