find the length of the following curve. if you have a grapher, you may want to graph the curve to see what…

Explanation:

Step1: Recall arc - length formula

The arc - length formula for a function $y = f(x)$ from $x=a$ to $x = b$ is $L=\int_{a}^{b}\sqrt{1+(y')^{2}}dx$. First, find the derivative of $y=\frac{1}{27}(9x^{2}+6)^{\frac{3}{2}}$. Using the chain - rule, if $u = 9x^{2}+6$, then $y=\frac{1}{27}u^{\frac{3}{2}}$. The derivative of $y$ with respect to $u$ is $\frac{1}{27}\times\frac{3}{2}u^{\frac{1}{2}}=\frac{1}{18}u^{\frac{1}{2}}$, and the derivative of $u$ with respect to $x$ is $18x$. So, $y'=\frac{1}{18}(9x^{2}+6)^{\frac{1}{2}}\times18x=x\sqrt{9x^{2}+6}$.

Step2: Calculate $(y')^{2}$

$(y')^{2}=x^{2}(9x^{2}+6)=9x^{4}+6x^{2}$.

Step3: Calculate $1+(y')^{2}$

$1+(y')^{2}=1 + 9x^{4}+6x^{2}=(3x^{2}+1)^{2}$.

Step4: Calculate the arc - length integral

$L=\int_{0}^{6}\sqrt{(3x^{2}+1)^{2}}dx=\int_{0}^{6}(3x^{2}+1)dx$. Integrating term - by - term, $\int(3x^{2}+1)dx=x^{3}+x+C$.

Step5: Evaluate the definite integral

$L=(x^{3}+x)\big|_{0}^{6}=(6^{3}+6)-(0^{3}+0)=216 + 6=222$.

Answer:

$222$