Question

In Exercises 29-32, find the volume of the solid described. Find the volume of the solid generated by revolving the region bounded by \(y=\sqrt{x}\) and the lines \(y=2\) and \(x=0\) about (a) the \(x\) -axis\( (b) the \)y\( -axis.\) (c) the line \(y=2\) (d) the line \(x=4\)

Short answer

The volumes of the solid when rotated about the x-axis, y-axis, y = 2, and x = 4 are \(8\pi\), \(8\pi\), \(0\), and \(0\), respectively.

Step-by-step solution

Identify the function and the interval

The solid is generated by revolving the region bounded by \(y=\sqrt{x}\), \(y = 0\), \(y = 2\), and \(x = 0\). These bounds form a region on the \(xy\)-plane, and our job is to measure the volumes of solids generated by rotating this region about different axes.

Volume about the x-axis

The volume \(V\) when the region is revolved around the \(x\)-axis (y=0) is given by the equation \(V = \pi \int_{a} ^{b} [R(x)]^2 dx\), where \(R(x)\) is the radius at a given \(x\). Notice that \(R(x) = y = \sqrt{x}\). The limits of integration are \(a = 0\) and \(b = 4 (y^2)\).\nSo, \(V = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx = \pi [x^2 / 2 |_0^4 = \pi (8) = 8\pi\)

Volume about the y-axis

We switch to use \(x\) as function of \(y\) that is \(x = y^2\), then for our region, the interval on y that is from \(y = 0\) to \(y = 2\). The volume when the region is revolved around the \(y\)-axis (x=0) is given by the equation \(V = 2\pi \int_{a} ^{b} y * R(y) dy\), where \(R(y) = x = y^2\). So\n\(V = 2\pi \int_{0}^{2} y^3 dy = 2\pi [y^4 / 4 |_0^2 = 2\pi (4) = 8\pi\)

Volume about the line \(y = 2\)

When revolving around the line \(y = 2\), the radius changes. \(R(x) = 2 - \sqrt{x}\). The volume \(V\) is found using the formula \(V = \pi \int_{a}^{b} [R(x)]^2 dx\), so \(V = \pi \int_{0}^{4} (2 - \sqrt{x})^2 dx = \pi \int_{0}^{4} (4 - 4\sqrt{x} + x) dx = \pi [4x - 8\sqrt{x}*2/3 + x^2/2 |_0^4 = \pi (0) = 0\)

Volume about the line \(x = 4\)

We will use \(x\) as function of \(y\), the interval on y is again from \(y = 0\) to \(y = 2\). As the axis of revolution has moved to \(x = 4\), the radius of revolution will now be \(R(y) = 4 - y^2\). The volume is given by \(V = 2\pi \int_{a}^{b} y * R(y) dy = 2\pi \int_{0}^{2} y * (4 - y^2) dy = 2\pi [(4y^2/2 - y^4/4) |_0^2 = 2\pi[(4*2/2 - 4) - 0 ] = 2\pi[4 - 4] = 0 \)