Q. 0 - Calculus

Question

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

Step-by-Step Solution

Answer

• $\text{Arc length}={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx$.
• 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 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 $\text{Arc length}={\int }_a^b\sqrt{1+{\left(f\text{'}\left(x\right)\right)}^2}dx$

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

  $S=2\pi rs$ ,

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

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