Section 7.2: Problem 13 (1 point) Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. \[ \begin{array}{l} y=\frac{1}{x^{4}}, y=0, x=2, x=5 \\ \text { about } y=-3 \end{array} \]

Step 1 ：The volume of a solid obtained by rotating a region bounded by curves about a specified axis can be found using the disk method or the washer method. In this case, we will use the washer method.

Step 2 ：The washer method formula is: \(V = \pi \int [R(x)]² - [r(x)]² dx\) where \(R(x)\) is the outer radius and \(r(x)\) is the inner radius.

Step 3 ：The outer radius \(R(x)\) is the distance from the axis of rotation to the farthest boundary curve, and the inner radius \(r(x)\) is the distance from the axis of rotation to the nearest boundary curve.

Step 4 ：In this case, the axis of rotation is \(y = -3\), the outer radius \(R(x)\) is the distance from \(y = -3\) to \(y = 0\), which is 3, and the inner radius \(r(x)\) is the distance from \(y = -3\) to \(y = 1/x⁴\), which is \(3 + 1/x⁴\).

Step 5 ：So, the volume \(V\) of the solid is: \(V = \pi \int_{2}^{5} [R(x)]² - [r(x)]² dx = \pi \int_{2}^{5} [3]² - [3 + 1/x⁴]² dx = \pi \int_{2}^{5} 9 - (9 + 6/x⁴ + 1/x⁸) dx = \pi \int_{2}^{5} -6/x⁴ - 1/x⁸ dx\)

Step 6 ：Now, we can integrate: \(= \pi [-6 \int_{2}^{5} x^-4 dx - \int_{2}^{5} x^-8 dx] = \pi [1.5x^-3 |_{2}^{5} - 0.125x^-7 |_{2}^{5}] = \pi [1.5(5^-3 - 2^-3) - 0.125(5^-7 - 2^-7)] = \pi [1.5(0.008 - 0.125) - 0.125(0.00000032 - 0.00000078125)] = \pi [1.5(-0.117) - 0.125(-0.00000046125)] = \pi [-0.1755 + 0.00000005765625] = \pi [-0.17544234375]\)

Step 7 ：Since volume cannot be negative, we take the absolute value: \(V = \pi * 0.17544234375\)

Step 8 ：So, the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis is approximately 0.5511 cubic units.

Step 9 ：\(\boxed{V \approx 0.5511 \text{ cubic units}}\)