Question

The region bounded by \(y=\sqrt{x}, y=0, x=0,\) and \(x=4\) is revolved about the \(x\) -axis. (a) Find the value of \(x\) in the interval [0,4] that divides the solid into two parts of equal volume. (b) Find the values of \(x\) in the interval [0,4] that divide the solid into three parts of equal volume.

Short answer

The value of \(x\) that would divide the volume into two equal parts is \(x=2\). To divide the solid into three equal volumes, the two values required are \(x = 1.587\) and \(x = 3.174\)

Step-by-step solution

Setting up the basic equation

First step is to derive the volume of a solid of revolution formula. For a region bounded by a curve and a line of rotation, volume is given by \(V=\pi \int_a^b[f(x)]^2 dx\). In this case, it is the region under the curve \(y=\sqrt{x}\) between \(x=0\) and \(x=4\) revolved about the \(x\) -axis, thus, \(f(x) = \sqrt{x}\) and the volume becomes: \(V=\pi \int_0^4 x dx \)

Calculating total volume

Calculate the total volume of the solid by evaluating this integral. Simplify the equation to: \(V= \pi [\frac{x^2}{2}]_0^4\). This calculation gives the total volume as: \(V = 8\pi cubic\space units\).

Part A: Locate x that splits the volume in two equal parts

The volume should be split into two equal parts, that is each part should have a volume of \(4\pi\). So set up the integral equation: \( \pi \int_0^x t dt = 4\pi\). Solving this equation for \(x\), we get \(x=2\).

Part B: Locate x-values that divide the volume into three equal parts

The volume should be split into three equal parts, each with a volume of \(\frac{8\pi}{3}\). For the first \(x\)-value that divides the solid volume in three, set up the integral equation: \( \pi \int_0^x t dt = \frac{8\pi}{3}\). Solve this for \(x\) to get the first \(x\)-value. Repeat the process for the second \(x\)-value: \( \pi \int_x^4 t dt = \frac{8\pi}{3}\), then solve for \(x\).