Find the volume of the solid formed by rotating the region enclosed by
\[
y=e^{4 x}+6, y=0, x=0, x=0.3
\]
about the $x$-axis. Enter your answer as an approximation, accurate up to three or more decimal places.
\[
V=
\]
Answer
Final Answer: The volume of the solid formed by rotating the region enclosed by \(y=e^{4 x}+6\), \(y=0\), \(x=0\), \(x=0.3\) about the \(x\)-axis is approximately \(\boxed{18.849}\) cubic units.
Step 1 ：The volume of the solid formed by rotating a curve \(y=f(x)\) from \(x=a\) to \(x=b\) about the x-axis is given by the formula: \(V = \pi \int_{a}^{b} [f(x)]^2 dx\).
Step 2 ：In this case, \(f(x) = e^{4x} + 6\), \(a=0\), and \(b=0.3\). We can use this formula to calculate the volume.
Step 3 ：Substitute the values of \(f(x)\), \(a\), and \(b\) into the formula.
Step 4 ：Calculate the integral to find the volume.
Step 5 ：Final Answer: The volume of the solid formed by rotating the region enclosed by \(y=e^{4 x}+6\), \(y=0\), \(x=0\), \(x=0.3\) about the \(x\)-axis is approximately \(\boxed{18.849}\) cubic units.
