Find the volume of the solid obtained by rotating the region bounded by the given curves about the line $x=2$ \[ x=y^{6}, x=1 \] Answer:

Step 1 ：The volume of the solid obtained by rotating the region bounded by the curves \(x=y^6\) and \(x=1\) about the line \(x=2\) can be found using the method of cylindrical shells.

Step 2 ：The formula for the volume of a solid of revolution using cylindrical shells is given by \[V = 2\pi \int_{a}^{b} r(x)h(x) dx\] where \(r(x)\) is the radius of the shell and \(h(x)\) is the height of the shell.

Step 3 ：In this case, the radius of the shell is \(2-x\) and the height of the shell is \(y\), which is equal to \(x^{1/6}\).

Step 4 ：So, the volume of the solid is given by \[V = 2\pi \int_{0}^{1} (2-x)x^{1/6} dx\]

Step 5 ：To solve this integral, we can use the power rule for integration, which states that the integral of \(x^n\) with respect to \(x\) is \(\frac{1}{n+1}x^{n+1}\).

Step 6 ：So, we get \[V = 2\pi \left[ \int_{0}^{1} 2x^{1/6} dx - \int_{0}^{1} x^{7/6} dx \right]\]

Step 7 ：This simplifies to \[V = 2\pi \left[ \frac{6}{7}x^{7/6} - \frac{6}{13}x^{13/6} \right]_{0}^{1}\]

Step 8 ：Substituting the limits of integration, we get \[V = 2\pi \left[ \frac{6}{7} - \frac{6}{13} \right]\]

Step 9 ：Simplifying the fractions, we get \[V = 2\pi \left[ \frac{78-42}{91} \right]\]

Step 10 ：Further simplifying, we get \[V = 2\pi \left[ \frac{36}{91} \right]\]

Step 11 ：So, the volume of the solid is \(\boxed{\frac{72\pi}{91}}\) cubic units.