Answer
vertical plane when $\theta=\dfrac{\pi}{4}$
Work Step by Step
In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$
Conversion of rectangular to cylindrical coordinate system gives: $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
Here, we have $\theta=\dfrac{\pi}{4}$
We know that $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$
Thus, we have
$\sqrt{x^2+y^2}=2$
This can be written as:
$x^2+y^2=2^2$
Thus, it represents an equation of a vertical plane when $\theta=\dfrac{\pi}{4}$
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