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.)

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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:

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

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.)

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