Problem 3 - Calculus Early Transcendentals

Question

The region bounded by the curves \(y=2 x\) and \(y=x^{2}\) is revolved about the \(x\) -axis. Give an integral for the volume of the solid that is generated.

Step-by-Step Solution

Answer

Answer: \(V=2\pi\int_{0}^{4} y\left(\sqrt{y} - \dfrac{y}{2}\right) ~dy\)

Step 1: Find Intersection Points of the Curves

To find the intersection points of the curves \(y=2x\) and \(y=x^2\), we need to set them equal to each other and solve for x: \(2x = x^2 \Rightarrow x^2 - 2x = 0\) Now, factor the equation: \(x(x-2)=0\) So, there are two solutions: \(x=0\) and \(x=2\). The corresponding y values can be found by plugging these values into either of the two equations. For simplicity, we'll use the first equation, \(y=2x\): - For \(x=0\): \(y=2(0)=0\) - For \(x=2\): \(y=2(2)=4\) The intersection points are thus \((0,0)\) and \((2,4)\).

Step 2: Set Up the Integral for the Volume using Cylindrical Shells

We're going to use the method of cylindrical shells to find the volume. The formula for the volume of a solid generated by revolving a region in the x-y plane about the x-axis is: \(V = 2\pi\int_{a}^{b} y \cdot R(y) ~dy\) Where a and b are the y values of the intersection points and \(R(y)\) is the radius of the cylindrical shell at a particular y value, which will be the difference between the x-values of the two functions at that y value. We need to rewrite both equations in terms of y: 1. \(y=2x \Rightarrow x=\dfrac{y}{2}\) 2. \(y=x^2 \Rightarrow x=\sqrt{y}\) Now find the difference between the x-values, which will be our radius function \(R(y)\): \(R(y) = \sqrt{y} - \dfrac{y}{2}\) Now we can set up the integral: \(V=2\pi\int_{0}^{4} y\left(\sqrt{y} - \dfrac{y}{2}\right) ~dy\) This integral represents the volume of the solid generated by revolving the region bounded by the curves \(y=2x\) and \(y=x^2\) about the x-axis.