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nvert the equation \( x^{2}+y^{2}=49 \) to spherical coordinates and write in standard form, solving for \( \rho \). se rho, theta, phi because we don't have those characters on our keyboard.)

Ask by Hanson Patel. in the United States
Mar 08,2025

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In spherical coordinates, the equation \( x^{2} + y^{2} = 49 \) becomes \( \rho = \frac{7}{\sin \phi} \).

Solution

To convert the equation \( x^{2} + y^{2} = 49 \) to spherical coordinates, we start by recalling the relationships between Cartesian coordinates \((x, y, z)\) and spherical coordinates \((\rho, \theta, \phi)\): - \( x = \rho \sin \phi \cos \theta \) - \( y = \rho \sin \phi \sin \theta \) - \( z = \rho \cos \phi \) Given the equation \( x^{2} + y^{2} = 49 \), we can substitute the expressions for \(x\) and \(y\) in terms of spherical coordinates: \[ x^{2} + y^{2} = (\rho \sin \phi \cos \theta)^{2} + (\rho \sin \phi \sin \theta)^{2} \] Now, we can simplify this expression: \[ x^{2} + y^{2} = \rho^{2} \sin^{2} \phi \cos^{2} \theta + \rho^{2} \sin^{2} \phi \sin^{2} \theta \] Factoring out \(\rho^{2} \sin^{2} \phi\): \[ x^{2} + y^{2} = \rho^{2} \sin^{2} \phi (\cos^{2} \theta + \sin^{2} \theta) \] Using the Pythagorean identity \(\cos^{2} \theta + \sin^{2} \theta = 1\), we have: \[ x^{2} + y^{2} = \rho^{2} \sin^{2} \phi \] Now, we can set this equal to 49: \[ \rho^{2} \sin^{2} \phi = 49 \] To solve for \(\rho\), we rearrange the equation: \[ \rho^{2} = \frac{49}{\sin^{2} \phi} \] Taking the square root of both sides gives us: \[ \rho = \frac{7}{\sin \phi} \] Thus, the equation \( x^{2} + y^{2} = 49 \) in spherical coordinates, solved for \(\rho\), is: \[ \rho = \frac{7}{\sin \phi} \]

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To convert the equation \( x^{2}+y^{2}=49 \) into spherical coordinates, we use the relationships \( x = \rho \sin \phi \cos \theta \) and \( y = \rho \sin \phi \sin \theta \). Substituting these into the equation gives us: \[ (\rho \sin \phi \cos \theta)^{2} + (\rho \sin \phi \sin \theta)^{2} = 49 \] This simplifies to: \[ \rho^{2} \sin^{2} \phi (\cos^{2} \theta + \sin^{2} \theta) = 49 \] Using the identity \( \cos^{2} \theta + \sin^{2} \theta = 1 \), we find: \[ \rho^{2} \sin^{2} \phi = 49 \] Solving for \( \rho \) yields: \[ \rho = \frac{7}{\sin \phi} \] This equation represents a cylinder-like surface in spherical coordinates. In spherical coordinates, this represents a vertical cylinder with a radius of 7, extending indefinitely along the z-axis. While you may imagine it as a simple shape, in three dimensions, the spatial relationships can become quite complex! The angle \( \phi \) essentially determines how "high" or "low" the point is while \( \rho \) will dictate how far out it reaches from the center depending on the angle.

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