# Areas and lengths in polar coodinates

## Areas in polar coordinates.

• Consider a polar region $\mathscr R:0\le r\le f(\theta),\,\alpha\le\theta\le\beta.$
• For this to make sense, we need the interval $$[\alpha,\beta]$$ not to cover more than a full period: $\beta-\alpha\le2\pi.$
• We will calculate the area of $$\mathscr R$$ in a way similar to how we did this in Cartesian coordinates.
• For each $$n>0$$, we slice the parameter interval $$[\alpha,\beta]$$ to $$n$$ subintervals $$[\theta_{i-1},\theta_i]$$ of equal length $$\Delta\theta=\frac{\beta-\alpha}{n}$$: $\theta_0=\alpha,\,\theta_i=\alpha+\Delta\theta,\dotsc,\theta_i=\alpha+i\Delta\theta,\dotsc,\theta_n=\alpha+n\Delta\theta=\beta.$
• In each subinterval, we choose a sample point $$\theta_{i-1}\le\theta_i^*\le\theta_i$$.
• We add up the areas of the circle sectors with radius $$f(\theta_i^*)$$ and angle $$\Delta\theta$$, thus getting the partial sum $$R_n=\sum_{i=1}^n\frac12f(\theta_i^*)^2\Delta\theta$$
• Then the area will be $A(\mathscr R)=\lim_{n\to\infty}R_n=\int_\alpha^\beta\frac12f(\theta)^2\,\mathrm d\theta.$
• Similarly, if the polar region is between two polar curves: $\mathscr R:f(\theta)\le r\le g(\theta),\,\alpha\le\theta\le\beta,$ then the area is: $A(\mathscr R)=\int_\alpha^\beta\frac12(g(\theta)^2-f(\theta)^2)\,\mathrm d\theta.$
• In the latter case, we might have to find the intersections between the polar curves $r=f(\theta)\text{ and }r=g(\theta).$
• Since we have $$(r,\theta)=(-r,\theta+\pi)$$, in the general case, you have to find the solutions of two equations $f(\theta)=g(\theta)\text{ and }f(\theta)=-g(\theta+\pi).$
• If you draw the curves and understand what's going on, then you might be able to avoid having to solve both equations.
• Example. Let's find the intersection points of the polar curves $r=\cos2\theta\text{ and }r=\frac12.$
• First, let's solve $$\cos2\theta=\frac12$$. We get $$\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$$.
• Now for $$\cos2\theta=-\frac12$$. This gives $$\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$$.
• Note that in this case, we need all of these, since there are 8 intersection points.
• Example. Let's calculate the area enclosed by one loop of the four-petal rose $$r=\cos2\theta$$.
• Note we get one loop for example on the interval $$-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$$.
• But by symmetry, it will be enough to calculate the area enclosed by a half petal, for example the one with $$0\le\theta\le\pi/4$$.
• Now we can use the formula: $A(\mathscr R)=2\int_0^{\pi/4}\frac12\cos^22\theta\,\mathrm d\theta=\frac12\int_0^{\pi/4}1+\cos4\theta\,\mathrm d\theta=\frac{\pi}{8}.$
• Example. Let's find the area of the region inside the circle $$r=3\sin\theta$$ and outside the cardioid $$r=1+\sin\theta$$.
• First of all, we need to find for which $$\theta$$ do we get $$1+\sin\theta\le3\sin\theta$$.
• Since we get all of the circle when $$0\le\theta\le\pi$$, and in that interval $$3\sin\theta\ge0$$ and $$1+\sin\theta\ge0$$, in this case we don't have to check for intersections when one of the distance functions is negative.
• Therefore, the solutions of $$\frac12=\sin\theta$$ give all the intersection points: $$\theta=\frac{\pi}{6},\frac{5\pi}{6}$$.
• We can see on a sketch or by checking that $$3=3\sin\frac{\pi}{2}>1+\sin\frac{\pi}{2}=2$$ that we have $$3\sin\theta\ge1+\sin\theta$$ when $$\frac{\pi}{6}\le\theta\le\frac{5\pi}{6}$$.
• Therefore, the area is $A(\mathscr R)=\int_{\pi/6}^{5\pi/6}\frac12((3\sin\theta)^2-(1+\sin\theta)^2)\,\mathrm d\theta=\frac12\int_{\pi/6}^{5\pi/6}8\sin^2\theta-2\sin\theta-1\,\mathrm d\theta=\frac12\int_{\pi/6}^{5\pi/6}3-4\cos2\theta\,\mathrm d\theta+\sqrt3=\pi.$
• Exercises. 10.4: 5, 7, 17, 23, 27, 29, 31, 33.

## Lengths in polar coordinates

• We have seen last time that a polar curve $$C:r=f(\theta),\,\alpha\le\theta\le\beta$$ can be viewed as a parametric curve: $C:x=f(\theta)\cos\theta),\quad y=f(\theta)\sin\theta,\quad\alpha\le\theta\le\beta.$
• Correspondingly, its arc length can be calculated using the arc length of a parametric curve formula: $L(C)=\int_\alpha^\beta\sqrt{\left(\frac{\mathrm dx}{\mathrm d\theta}\right)^2+\left(\frac{\mathrm dy}{\mathrm d\theta}\right)^2}\,\mathrm d\theta.$
• Now since this is a special case, one can simplify the formula: $\left(\frac{\mathrm dx}{\mathrm d\theta}\right)^2+\left(\frac{\mathrm dy}{\mathrm d\theta}\right)^2=(f'(\theta)\cos\theta-f(\theta)\sin\theta)^2+(f'(\theta)\sin\theta+f(\theta)\cos\theta)^2=(f'(\theta))^2+(f(\theta))^2$ gives $L(C)=\int_\alpha^\beta\sqrt{(f(\theta))^2+(f'(\theta))^2}\,\mathrm d\theta$
• Example. Let's calculate the length of the cardioid $$r=1+\sin\theta$$.
• Using the formula, we get $L=\int_0^{2\pi}\sqrt{(1+\sin\theta)^2+\cos^2\theta}\,\mathrm d\theta=\int_0^{2\pi}\sqrt{2+2\sin\theta}\,\mathrm d\theta.$
• Exercises. 10.4: 45, 47.