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Find an equation in cylindrical coordinates for the rectangular equation \( y^{2}=10-z^{2} \) Entry Tip \#1: To enter \( \theta \), type "theta". Example: \( r^{*} \sin ( \) theta \( )=r \sin (\theta) \) Entry Tip \#2: To enter a squared trig function, wrap the outside of the trig function with parentheses. Example: \( (\sin (\text { theta }))^{\wedge} 2=\sin ^{2}(\theta) \)

Ask by Sherman French. in the United States
Mar 08,2025

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Answer

In cylindrical coordinates, the equation \( y^{2} = 10 - z^{2} \) becomes \( r^{2} \sin^{2}(\theta) = 10 - z^{2} \).

Solution

1. In cylindrical coordinates, we have the conversion \( y = r \sin (\theta) \). 2. Substitute this into the rectangular equation: \[ (r \sin (\theta))^{2} = 10 - z^{2}. \] 3. Simplify the left-hand side: \[ r^{2} \sin^{2} (\theta) = 10 - z^{2}. \] Thus, the equation in cylindrical coordinates is: \[ r^{2} \sin^{2}(\theta) = 10 - z^{2}. \]

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In cylindrical coordinates, the relationship between rectangular and cylindrical coordinates gives us \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( z = z \). Given the equation \( y^{2}=10-z^{2} \), we can convert it to cylindrical coordinates by substituting \( y \) with \( r \sin(\theta) \). This results in: \[ (r \sin(\theta))^{\wedge} 2 = 10 - z^{2} \] Now, rearranging gives: \[ r^{\wedge} 2 \sin^{\wedge} 2(\theta) + z^{\wedge} 2 = 10 \] This is the equation in cylindrical coordinates. To visualize this equation, you might think of it as a sort of "hyperbolic cylinder" that opens along the z-axis, where the cross-sections in the z-plane are hyperbolas. The interaction of z-values and the radial distance from the z-axis, determined by \( r \) and \( \theta \), defines a beautiful 3D shape!

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