**How to sketch image under function on given set**without wasting too much if your time.

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For the function, $f(r,theta)= (rcostheta , rsintheta )$, 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

$$f(r,0)=(r,0)$$

$$f(r,pi)=(-pi,0)$$

$$f(1,theta)=(costheta, sintheta)$$

$$f(2,0)=(2costheta, 2sintheta)$$

Thanks in advance

## Answer

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 $rin[1,2]$ this is the line segment between the points $(1,0)$ and $(2,0)$.

Your second values are wrong. They should be

$$f(r, pi)=(rcospi, rsinpi)=(rcdot -1, rcdot 0) = (-r, 0)$$

Again, $rin[1,2]$, so this is the line segment between the points $(-1,0)$ and $(-2,0)$.

Your third values are correct, $f(1, theta)=(costheta, sintheta)$. 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: $f(2, theta)=(2costheta, 2sintheta)$. 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.

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