find the lateral (side) surface area of the cone generated by revolving the line - segment y =…

Explanation:

Step1: Recall surface - area formula for revolution

The formula for the lateral surface area of a solid of revolution about the (x) - axis is (S = 2\pi\int_{a}^{b}y\sqrt{1+(y')^{2}}dx). Given (y=\frac{9}{7}x), first find (y'). Since (y = \frac{9}{7}x), then (y'=\frac{9}{7}).

Step2: Substitute (y) and (y') into the formula

Substitute (y=\frac{9}{7}x) and (y'=\frac{9}{7}) into (S = 2\pi\int_{a}^{b}y\sqrt{1+(y')^{2}}dx). Here, (a = 0) and (b = 7). [ \begin{align*} S&=2\pi\int_{0}^{7}\frac{9}{7}x\sqrt{1 + (\frac{9}{7})^{2}}dx\ &=2\pi\sqrt{1+\frac{81}{49}}\frac{9}{7}\int_{0}^{7}x dx \end{align*} ] Since (1+\frac{81}{49}=\frac{49 + 81}{49}=\frac{130}{49}), then (\sqrt{1+\frac{81}{49}}=\frac{\sqrt{130}}{7}). [ \begin{align*} S&=2\pi\times\frac{\sqrt{130}}{7}\times\frac{9}{7}\int_{0}^{7}x dx\ &=\frac{18\pi\sqrt{130}}{49}\int_{0}^{7}x dx \end{align*} ]

Step3: Evaluate the integral

We know that (\int_{0}^{7}x dx=\left[\frac{1}{2}x^{2}\right]{0}^{7}=\frac{49}{2}). [ \begin{align*} S&=\frac{18\pi\sqrt{130}}{49}\times\frac{49}{2}\ &=9\pi\sqrt{130} \end{align*} ] Let's check with the geometric formula. The base - radius (r) when (x = 7) is (y(7)=\frac{9}{7}\times7 = 9), the base - circumference (C = 2\pi r=18\pi), and the slant - height (l=\sqrt{7^{2}+9^{2}}=\sqrt{49 + 81}=\sqrt{130}). Using the geometric formula (S=\frac{1}{2}\times C\times l=\frac{1}{2}\times18\pi\times\sqrt{130}=9\pi\sqrt{130}). The integral that gives the surface area of the cone is (S = 2\pi\int{0}^{7}\frac{9}{7}x\sqrt{1+\left(\frac{9}{7}\right)^{2}}dx=\frac{18\pi\sqrt{130}}{49}\int_{0}^{7}x dx).

Answer:

(S = 2\pi\int_{0}^{7}\frac{9}{7}x\sqrt{1+\left(\frac{9}{7}\right)^{2}}dx)