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) = x^{3/2} from 0 ⇐> 4 $\begin{array}{rl}L& ={\int }_0^4\sqrt{1+(\frac{3}{2}{x}^{\frac{1}{2}}{)}^2}\\ & ={\int }_0^4(1+(\frac{9}{4}x){)}^{\frac{1}{2}}\\ & =\frac{2}{3}\cdot \frac{4}{9}\cdot (1+\frac{9}{4}x{)}^{\frac{3}{2}}\end{array}$ …

I’m not entirely sure why the book puts this at $\frac{2}{3}\frac{4}{9}(1+\frac{9}{4}x{)}^{\frac{3}{2}}{|}_0^4$.

• DONE Completed: 2022-06-01

I don’t know where the 4/9 comes from. I know it’s probably something to do with the chain rule, but not clear on the specifics.

So it seems like they’re doing a u-substitution of u=1+9/4x, but this means du = 9/4 dx. To make that equal out.. we have to multiply the whole thing by 4/9th.

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: $f(x)=\sqrt{r^2-x^2}$