Problem 4 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

a) 0 and 2 b) 0 and 1 c) 1 and 2 d) 1 and 3 Correct Answer: a) 0 and 2

Step 1: Find the points of intersection of the two curves

To find the points of intersection, we need to solve the equation \(2x = x^2\): \begin{align*} 2x &= x^2 \\ x^2 - 2x &= 0 \\ x(x - 2) &= 0 \end{align*} The two points of intersection are \(x = 0\) and \(x = 2\).

Step 2: Set up the integral for the volume of the solid using the shell method

The radius of the cylindrical shell is \(x\), and its height is the difference between the functions \(y = x^2\) and \(y = 2x\). Therefore, we have the following formula for the volume of the solid generated: \begin{align*} V = 2 \pi \int_a^b x(y_{out} - y_{in}) dx \end{align*} Here, \(a\) and \(b\) are the points of intersection (0 and 2). \(y_{out}\) refers to the outer curve, while \(y_{in}\) refers to the inner curve. Since we're revolving around the y-axis, the outer curve is \(y = 2x\) and the inner curve is \(y = x^2\). We then have the following integral for the volume: \begin{align*} V = 2 \pi \int_0^2 x(2x - x^2) dx \end{align*}

Step 3: Final Answer

The integral for the volume of the solid generated when the region bounded by the curves \(y = 2x\) and \(y = x^2\) is revolved around the y-axis is: \begin{align*} V = 2 \pi \int_0^2 x(2x - x^2) dx \end{align*}