6.4 Exercises

31. An alternative derivation of the surface area formula

Assume $f$ is smooth on $[a,b]$ in the usual way. In the $k$th subinterval $[x_{k-1},x_{k}]$, construct the tangent line to the curve at the midpoint $m_k=(x_{k-1}+x_{k})/2$, as in the accompanying figure.

 

$$r_1 +r_2 =2f(m_k)$$

If line $l$ is the tangent Line at point $(m_k, f(m_k))$, then

$$l: y-f(m_k)=f^{\prime}(m_k)(x-m_k)$$

$(x_{k-1},r_1 ), (x_k, r_2)$ are put on $l$

$$r_1 -f(m_k)=f^{\prime}(m_k)(x_{k-1}-m_k)-------(1)$$

$$r_2 -f(m_k)=f^{\prime}(m_k)(x_{k}-m_k)----------(2)$$

$(1)-(2)$

$$r_1 -r_2=f^{\prime}(m_k)(x_{k-1}-x_k)=f^{\prime}(m_k)\Delta x_{k}$$

$$\therefore\;\;r_1 =f(m_k)-f^{\prime}(m_k)\frac{\Delta x_{k}}{2}, \;\;r_2 =f(m_k)+f^{\prime}(m_k)\frac{\Delta x_{k}}{2}$$

The lateral surface area of the cone swept out by the tangent line segment as it revolves about the x-axis is

$$S_k=2\pi f(m_k) \sqrt{(\Delta x_k)^2 +(\Delta y_k)^2}=2\pi f(m_k) \sqrt{1+(f^{\prime}(m_k))^2} \Delta x_k$$

$$S\approx \sum_{k=1}^{n} S_k =\sum_{k=1}^{n}  2\pi f(m_k) \sqrt{1+(f^{\prime}(m_k))^2} \Delta x_k$$

$$\therefore \;\;S=\lim_{n\rightarrow \infty}\sum_{k=1}^{n}  2\pi f(m_k) \sqrt{1+(f^{\prime}(m_k))^2} \Delta x_k=\int_{a}^{b} 2\pi f(x)\sqrt{1+(f^{\prime}(x))^2}dx$$