Question

consider a circular cone of radius 3 and height 5, 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.

Explanation:

Step1: Find the linear - function formula

The linear function passes through the points $(0,0)$ and $(5,3)$. The slope $m$ of the line is given by $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{3 - 0}{5 - 0}=\frac{3}{5}$. Using the point - slope form $y - y_1=m(x - x_1)$ with the point $(0,0)$, the linear function is $y = f(x)=\frac{3}{5}x$.

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

The volume $V$ of a solid of revolution using the disk method when rotating about the $x$ - axis is $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $a = 0$, $b = 5$, and $f(x)=\frac{3}{5}x$. So $V=\pi\int_{0}^{5}(\frac{3}{5}x)^{2}dx=\pi\int_{0}^{5}\frac{9}{25}x^{2}dx$.

Step3: Integrate the function

We know that $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$). So, $\pi\int_{0}^{5}\frac{9}{25}x^{2}dx=\frac{9\pi}{25}\times\frac{x^{3}}{3}\big|_{0}^{5}$.

Step4: Evaluate the definite integral

$\frac{9\pi}{25}\times\frac{x^{3}}{3}\big|_{0}^{5}=\frac{3\pi}{25}x^{3}\big|_{0}^{5}=\frac{3\pi}{25}(5^{3}-0^{3})=\frac{3\pi}{25}\times125 = 15\pi$.

Answer:

The formula for the linear function is $y = f(x)=\frac{3}{5}x$ and the volume of the cone is $15\pi$.