is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Integrating Polar and Parametric Functions

Parametric Functions

MathJax TeX Test Page If you had this set of equations: $x = \cos{t}$, and $y = \sin{t}$, and you want to find the area of that ste of parametric equations. We could do $$\text{The area is} \int\mathrm{y}\, \mathrm{d}x$$ $$x = \cos{t}, dx = -\sin{t}dt \rightarrow \int\mathrm{\sin{t}*-\sin{t}}\, \mathrm{d}t = \int\mathrm{-\sin^2{t}}\, \mathrm{d}t$$ $$\cos{2t} = 1 - 2\sin^2{t} \rightarrow -\sin^2{t} = \frac{1}{2}(cos{2t} - 1)$$ $$=\frac{1}{2}\int\mathrm{\cos{2t} - 1}\, \mathrm{d}t = \frac{\sin{2t}}{4} - \frac{x}{2} + C$$ $$\text{If you graph the system, you get that it is a unit circle. So, you can integrate from 0 to 2}\pi$$ $$(\frac{\sin{2t}}{4} - \frac{x}{2})|^{2\pi}_{0} = -\pi$$ $$\text{Note that that is the integral, not the area. The area is the absolute value, which is } \pi$$

Polar Equations

MathJax TeX Test Page You can think about doing area under a polar function as the sum of many sectors. The area of a sector is $\frac{1}{2\pi}(\pi{}r^2) = \frac{r^2}{2}$, so the area of the entire part is $\frac{1}{2}\int_{a}^{b}\mathrm{r(\theta{})^2}\, \mathrm{d}\theta{}$


r = 1 - 2cos( θ)

r = 1 - 2cos(θ)

MathJax TeX Test Page Find the area of the outer loop of $r = 1 - \cos{\theta{}}$ For this problem, you have to take total area - the inner loop. First, you have to figure out that the inner loop goes from $-\frac{\pi}{3} \rightarrow \frac{\pi}{3}$ $$\text{The outer loop = } \frac{1}{2}\int_{\frac{\pi}{3}}^{\frac{5\pi{}}{3}}\mathrm{(1-\cos{\theta{}})^2}\, \mathrm{d}\theta{}$$ $$\text{The inner loop = } \frac{1}{2}\int_{\frac{-\pi{}}{3}}^{\frac{\pi}{3}}\mathrm{(1-cos{\theta})^2}\, \mathrm{d}\theta{}$$ To calculate the integral of $(1 - \cos^{x})^2$, you power reduce $$\int\mathrm{\cos^2{x} - 2\cos{x} + 1}\, \mathrm{d}x = \int\mathrm{\frac{\cos{2x}}{2} + \frac{1}{2} - 2\cos{x} + 1}\, \mathrm{d}x$$ $$=\frac{\sin{2x}}{4} + \frac{3}{2}x - 2\sin{x}$$ The outer loop - the inner loop = 4.60189
David Witten

Partial Fractions

Trigonometric Substitution