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.)

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: \[ \rho = \frac{7}{\sin \phi} \]