Question

Multiple Choice Let \(R\) be the region in the first quadrant bounded by the graph of \(y=8-x^{3 / 2}\) , the \(x\) -axis, and the \(y\) -axis. Which of the following gives the best approximation of the volume of the solid generated when \(R\) is revolved about the \(x\) -axis? (A) 60.3\(\quad\) (B) 115.2 (C) 225.4 (D) 319.7 (E) 361.9

Short answer

The best approximation of the volume of the solid generated when \(R\) is revolved about the \(x\)-axis is (B) 115.2.

Step-by-step solution

Setting Up the Integral

In the disk method, the volume V of a solid of revolution about the x-axis is computed as \(V = \pi \int_a^b [f(x)]^2 dx\), where \(f(x)\) is the function that describes the curve and [a, b] is the interval over which the curve is defined. For this problem, \(f(x) = 8 - x^{3/2}\), and the interval [a, b] is from 0 (the y-axis) to where the curve intersects the x-axis. To find this intersect, we would set \(y = 8 - x^{3/2} = 0\) and solve for \(x\). That results in \(x = 8^{2/3} = 4\).

Calculating the Integral

Substituting \(f(x)\) and [a, b] into the volume integral gives: \(V = \pi \int_0^{4} (8 - x^{3/2})^2 dx\). Now it’s a matter of calculus to solve this integral.

Evaluating the Integral and Selecting the Best Approximation

Evaluating \(V = \pi \int_0^{4} (8 - x^{3/2})^2 dx\) gives \(V = 256\pi / 5 = 160.84956\) - none of the given choices. However, as it’s asked for the best approximation, the closest choice is (B) 115.2