How to sketch image under function on given set For the function, $f(r,\theta)= (r\cos\theta , r\sin\theta )$, I want to sketch the image under $f$ of the set $S=[1,2]$ x$ [0,\pi]$
My first step was to find the images of $f$ along the borders of the line segments given by the rectangle.  However, I am unsure of how to proceed to find the sketch of the entire image.  Below is what I got from fixing $r$, and $\theta$ on the bordering line segments
$$f(r,0)=(r,0)$$
$$f(r,\pi)=(-\pi,0)$$
$$f(1,\theta)=(\cos\theta, \sin\theta)$$
$$f(2,0)=(2\cos\theta, 2\sin\theta)$$
Thanks in advance
 A: You have a good idea, but you made some mistakes in the implementation.
You have the first values correct: $f(r,0)=(r,0)$. Since $r\in[1,2]$ this is the line segment between the points $(1,0)$ and $(2,0)$.
Your second values are wrong. They should be
$$f(r,\ \pi)=(r\cos\pi,\ r\sin\pi)=(r\cdot -1,\ r\cdot 0) = (-r,\ 0)$$
Again, $r\in[1,2]$, so this is the line segment between the points $(-1,0)$ and $(-2,0)$.
Your third values are correct, $f(1,\ \theta)=(\cos\theta,\ \sin\theta)$. You should recognize this as a parametrization of the unit circle--remember your definitions for cosine and sine in trigonometry class? (Ask if you need more detail on this.) However, this is not the full circle, since the angle $\theta$ is limited to $[0,\ \pi]$. That gives the upper unit semicircle.
Your fourth values are almost correct: $f(2,\ \theta)=(2\cos\theta,\ 2\sin\theta)$. Again using trigonometry, this is the circle with its center at the origin and with radius $2$. Again, the limitations on the angle give the upper semicircle of radius $2$.
Putting those all together, the boundary of your region is the upper half of the "washer" with inner radius $1$ and outer radius $2$. The region is the boundary with its interior. Here is a graphic without shading the interior.

