Carefully sketch the graph of the polar curve r=theta^2, 0<=theta<=pie, and determine the exact area of the region bounded by this curve and the x axis. pleaseeee. Anybody.
Hello. Did you ever finish the previous question we discussed earlier today? Tell me, in a nutshell, what you currently understand about polar coordinates. What does "r" stand for? What does "theta" represent, and how is it measured? By "pie" I assume you mean the Greek character Pi. Have you tried graphing \[r = \theta^2?\]
you can sketch it via desmos. And the area is just a calc II problem
Desmos is a marvel, but I think it's important to know how to graph by hand when needed. @Khally92?
I haven't ever been asked to handsketch polar graphs, whether that's good or bad.
\(\color{#000000 }{ \displaystyle {\rm Area}=\frac{1}{2}\int_0^{\pi} \left(\theta ^2\right)^2 d\theta }\) that would be the set up, if I recall correctly.
well @mathmale I got the previous question figured out, I have no problem with the integration, but we weren't quite sure about the region. I have email my instructor for more clarification, thank you. and about this one
Sketch the point where \(\theta=\pi\) and where \(\theta=0\), and a couple of points between that to approximately visualize
Solomon, as usual your contributions are right on, mathematically correct, but I would much prefer that we let @khally92 do as much of the solution work as possible himself.
Sure:)
yeah @SolomonZelman mathhale is right.
welll r is the distance from the origin
You need to vary theta from 0 to Pi. why not choose a few values, such as pi/6, pi/3, etc.,, and find the corresponding radii r from the formula r = (theta)^2?
theta is the angle between the x axis and r
Right. r is the distance from the origin to a point on the curve. These points are determined by both the radius (r) and the angle (theta), just was you've seen in cartesian coordinates (x, y).
Mind making a table? For example: If theta = pi/6, r = (pi/6)^2.
You could round off your "r" to a single decimal point. All we need is a rough sketch of the figure generated as we let theta = 0 to theta=pi.
Thats what i am doing now
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