calculuspolar coordinates
Graph polar coordinates $ r=4\sin(3\theta) $
I was told by my teacher to split the graph into $3$ parts per quadrant and try those angles the problem arises when I plug $\dfrac{5\pi}{6}$ into the equation and I don't know what to do to solve it without a calculator. If someone has a better process to graphing polar equations/curves I would be grateful.
To integrate this region in polar coordinates, it is advisable to break up the integral into two parts, as shown in the figures below:
The two parts of the integral are divided by the diagonal line through the upper right corner of the rectangle. Since the sides of the rectangle are $a$ and $b$, this diagonal line is at the angle $\arctan \frac ba.$
For $0 \leq \theta \leq \arctan \frac ba,$ you would integrate over $0 \leq r \leq a \sec\theta,$ and for $\arctan \frac ba \leq \theta \leq \frac\pi2,$ you would integrate over $0 \leq r \leq b \csc\theta.$
If you actually try this, I think you'll find that it is no easier than doing the integration in rectangular coordinates. It may even be worse.
An alternative approach, rather than combining $x^2+y^2$ into $r^2$, is to integrate the terms separately:
$$\begin{eqnarray} m &=& \int_0^a\int_0^b (1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b dy\,dx +\int_0^a\int_0^b x^2 \,dy\,dx +\int_0^a\int_0^b y^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{x} &=& \int_0^a\int_0^b x(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b x \,dy\,dx +\int_0^a\int_0^b x^3 \,dy\,dx +\int_0^a\int_0^b xy^2 \,dy\,dx \end{eqnarray}$$ $$\begin{eqnarray} m\bar{y} &=& \int_0^a\int_0^b y(1+x^2+y^2)\,dy\,dx \\ &=& \int_0^a\int_0^b y \,dy\,dx +\int_0^a\int_0^b x^2y \,dy\,dx +\int_0^a\int_0^b y^3 \,dy\,dx \end{eqnarray}$$
Now you have nine integrals to solve, but they're all quite simple.
Suppose we have two curves that cross at right angles. We represent them in polar coordinates as functions $r_1$ and $r_2$ from angle to radius. Suppose the crossing happens at $(r, \theta)$, so that $r_1(\theta) = r_2(\theta) = r$.
For $i \in {1, 2}$, put $x_i(\theta) = r_i(\theta)\cos(\theta)$ and $y_i(\theta) = r_i(\theta)\sin(\theta)$. Then $\frac{dx_i}{d\theta} = \frac{dr_i}{d\theta}\cos(\theta) - r_i\sin(\theta)$, and $\frac{dy_i}{d\theta} = \frac{dr_i}{d\theta}\sin(\theta) + r_i\cos(\theta)$.
The curves cross at right angles, so we have $\frac{dx_1}{d\theta}\frac{dx_2}{d\theta} + \frac{dy_1}{d\theta}\frac{dy_2}{d\theta} = 0$. (This can be deduced from $\frac{dy_1}{dx_1} = -\frac{dx_2}{dy_2}$ when they are finite, but it is more general). Substitute:
$$ \left(\frac{dr_1}{d\theta}\cos(\theta)-r\sin(\theta)\right) \left(\frac{dr_2}{d\theta}\cos(\theta)-r\sin(\theta)\right) + \left(\frac{dr_1}{d\theta}\sin(\theta)+r\cos(\theta)\right) \left(\frac{dr_2}{d\theta}\sin(\theta)+r\cos(\theta)\right) = 0 $$
Multiply out the brackets, cancel the terms in $\cos(\theta)\sin(\theta)$, and use $\cos^2(\theta) + \sin^2(\theta) = 1$:
$$\frac{dr_1}{d\theta}\frac{dr_2}{d\theta} + r^2 = 0$$
Divide through by $r^2$ and subtract 1:
$$\left(\frac1r\frac{dr_1}{d\theta}\right) \left(\frac1r\frac{dr_2}{d\theta}\right) = -1$$
Best Answer
I'm going to answer your question in a more-general context- that way, you'll hopefully be able to sketch any (reasonable) polar curve.
Some useful points:
Tangents at the pole: these occur when $r=0$ (when the curve is at the origin).
In this case, tangents at the pole will be the values of $\theta$ satisfying $4\sin(3\theta)=0$ (which I'll leave you to solve).
Maxima- these are points which are furthest from the origin- maxima occur when $\frac{dr}{d\theta}=0$ (and $\frac{d^2r}{d\theta^2}<0$).
In this case, the maxima are (some) solutions to $12\cos(3\theta)=0$ (which, again, I'll leave you to solve).
I'll give you this one for your question- the period is $\frac{2\pi}{\color{green}{3}}$.
That means you can split the graph into $\color{green}{3}$ parts, each of which will be (rotationally) symmetric to the others- one third of the work!
Once you've found and plotted maxima and tangents at the pole, it is now simply a matter of connecting the dots.
The final result is this:
What I've left out are the values of $\theta$ for which $r$ is maximum (in blue), and the values of $\theta$ for which $r$ is minimum (red). Try and do this yourself!