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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