Question

In Exercises 35-38, use the cylindrical shell method to find the volume of the solid generated by revolving the region bounded by the curves about the y-axis. $$y=x^{2}, \quad y=2-x, \quad x=0, \quad \text { for } x \geq 0$$

Short answer

The volume of the solid generated is \(\frac{\pi}{2}\) cubic units.

Step-by-step solution

Intersection Point Finding

First, find the intersection of the two curves \(y=x^{2}\) and \(y=2-x\) by setting them equal to each other: \(x^{2} = 2 - x\). After solving this quadratic equation, two solutions are found: \(x = 1\) and \(x = 2\). Considering that \(x \geq 0\), these two intersections are valid.

Set-up the Integral for the Volume via Cylindrical Shells

The volume of the solid of revolution is found using the cylindrical shell method. The formula is \(V = 2\pi \int_{a}^{b} r(x) h(x) dx\), where \(r(x)\) is the radius and \(h(x)\) is the height of a cylindrical shell. In this problem, \(r(x) = x\) and \(h(x) = 2 - x - x^{2}\), and the limits of integration a and b are 0 and 1.

Compute the Integral

Substitute \(r(x)\) and \(h(x)\) into the formula and compute the integral. \(V = 2\pi \int_{0}^{1} x (2 - x - x^{2}) dx\). After calculation, the volume is \(V = \frac{\pi}{2}\) cubic units.