Question

(1 point) consider a circular cone of radius 3 and height 6, which we view horizontally as pictured below. our goal in this activity is to use a definite integral to determine the volume of the cone.
(a) find a formula for the linear function y = f(x) that is pictured above.
(b) for the representative slice of thickness δx that is located horizontally at a location x (somewhere between x = 0 and x = 6), what is the radius r at the representative slice? note that the radius depends on the value of x.
(c) what is the volume of the representative slice you found in (b)? (use d as the value for δx.)

Explanation:

Step1: Find the linear - function formula

We know that the cone has a radius \(r = 3\) at \(x = 6\) and \(r=0\) at \(x = 0\). Using the two - point form of a line \(y=mx + b\), where \(m=\frac{y_2 - y_1}{x_2 - x_1}\) and \(b\) is the \(y\) - intercept. Here, when \(x = 0,y = 0\) and when \(x = 6,y = 3\). The slope \(m=\frac{3-0}{6 - 0}=\frac{1}{2}\), and \(b = 0\). So the linear function \(y=f(x)=\frac{1}{2}x\).

Step2: Use the disk - method to find the volume of the cone

The volume \(V\) of a solid of revolution using the disk method is \(V=\pi\int_{a}^{b}[f(x)]^{2}dx\). Here, \(a = 0\), \(b = 6\), and \(f(x)=\frac{1}{2}x\). So \(V=\pi\int_{0}^{6}(\frac{1}{2}x)^{2}dx=\pi\int_{0}^{6}\frac{1}{4}x^{2}dx\).

Step3: Evaluate the integral

Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have \(\pi\int_{0}^{6}\frac{1}{4}x^{2}dx=\frac{\pi}{4}\times\frac{x^{3}}{3}\big|_{0}^{6}\).
\[

$$\begin{align*} \frac{\pi}{12}(6^{3}-0^{3})&=\frac{\pi}{12}\times216\\ &=18\pi \end{align*}$$

\]

Answer:

The volume of the cone is \(18\pi\) and the linear function \(y = f(x)=\frac{1}{2}x\)