Calculating arc length and surface area

calculus

Arc length for lines

So this is about calculating the length of the line on a curve, not the area underneath.

The formula is:

$$\begin{array}{r}L\approx \sum _{i=1}^n\sqrt{\Delta x_i^2+\Delta y_i^2}\end{array}$$

This is because we’re actually calculating the hypotenuse of a triangle with height $\Delta y$ and width $\Delta x$. This is only an approximation until we can get x or y isolated.

To do that isolation, we can pull out the dx^2 param.

$$\begin{array}{rl}& \sum _{i=1}^n\sqrt{\Delta x_i^2(1+\frac{\Delta y_i^2}{\Delta y_i^2})}\\ & \\ =& \sum _{i=1}^n\sqrt{1+\frac{\Delta y_i^2}{\Delta y_i^2}}\Delta x_i\end{array}$$

The Mean Value Theorem of Differentiation says that $f^{\prime }(c_i)=\frac{\Delta y}{\Delta x}$, so we can substitute that, which gives us:

$$\begin{array}{r}L={\int }_a^b\sqrt{1+f^{\prime }(x{)}^2}dx\end{array}$$

assuming f is differentiatable on [a,b] and f’ is continuous on [a,b]

Example

7.4.1

arc length of f(x) = x3/2 from 0 <=> 4

$$\begin{array}{rl}L& ={\int }_0^4\sqrt{1+(\frac{3}{2}{x}^{\frac{1}{2}}{)}^2}\\ & \\ & ={\int }_0^4(1+(\frac{9}{4}x){)}^{\frac{1}{2}}\\ & \\ & =\frac{2}{3}\cdot \frac{4}{9}\cdot (1+\frac{9}{4}x{)}^{\frac{3}{2}}\\ & \end{array}$$

…

Arc length for circles

circumfrence of a circle is:

$$\begin{array}{r}{r}^2=x^2+y^2\\ \\ y^2=r^2-x^2\\ \\ y=\pm \sqrt{r^2-x^2}\end{array}$$

for the top half: $f(x)=\sqrt{r^2-x^2}$