Question

Problem Zero: Read the section and make your own summary of the material.

Short answer

• $\text{Arclength}={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx$.
• The Surface area of a Frustum is $S=2\pi rs$, where, $r=\frac{p+q}{2}$is the average radius of the frustum.
• Surface area by a definite integral $S=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\sqrt{1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2}d\mathrm{x}$.

Step-by-step solution

Step 1. Given Information.

Arc length and surface area -

• Approximating arc length with a sum of line-segment lengths.
• Approximating surface area with a sum of frustum surface areas.
• Using definite integrals to calculate exact arc lengths and surface areas

Step 2. Approximating arc length with a sum of line-segment lengths.

The arc length of $f\left(x\right)$from $x=a$to $x=b$is given as follows $\text{Arclength}={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx$

Here, $f\left(x\right)$is a differentiable function with a continuous derivative.

Step 3. Approximating surface area with a sum of frustum surface areas.

A frustum with radius $p$and q and slant length s, Then the surface area is given by

role="math" localid="1650524759896" $S=2\pi rs$,

where $r=\frac{p+q}{2}$ is the average radius of the frustum.

Step 4. Using definite integrals to calculate exact arc lengths and surface areas.

If $f\left(x\right)$is a positive, differentiable function with a continuous derivative. The surface area of the solid of revolution obtained by revolving $f\left(x\right)$around the x-axis from $x=a$to $x=b$ is

role="math" localid="1650524941029" $S=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{f}\left(\mathrm{x}\right)\sqrt{1+{\left(\mathrm{f}\text{'}\left(\mathrm{x}\right)\right)}^2}d\mathrm{x}$.