The problem requires converting equations from rectangular coordinates $(x, y, z)$ to cylindrical coordinates $(r, \theta, z)$ or spherical coordinates $(\rho, \theta, \phi)$. Specifically, we need to solve the following problems: 17. Convert $x^2 + y^2 = 9$ to cylindrical coordinates. 18. Convert $x^2 - y^2 = 25$ to cylindrical coordinates. 19. Convert $x^2 + y^2 + 4z^2 = 10$ to cylindrical coordinates. 20. Convert $x^2 + y^2 + 4z^2 = 10$ to spherical coordinates. 21. Convert $2x^2 + 2y^2 - 4z^2 = 0$ to spherical coordinates.

1. Problem Description

The problem requires converting equations from rectangular coordinates $(x, y, z)$ to cylindrical coordinates $(r, \theta, z)$ or spherical coordinates $(\rho, \theta, \phi)$.

Specifically, we need to solve the following problems:

1

7. Convert $x^2 + y^2 = 9$ to cylindrical coordinates.

1

8. Convert $x^2 - y^2 = 25$ to cylindrical coordinates.

1

9. Convert $x^2 + y^2 + 4z^2 = 10$ to cylindrical coordinates.

2

0. Convert $x^2 + y^2 + 4z^2 = 10$ to spherical coordinates.

2

1. Convert $2x^2 + 2y^2 - 4z^2 = 0$ to spherical coordinates.

2. Solution Steps

Conversion Formulas:

Cylindrical Coordinates:

$x = r\cos\theta$

$y = r\sin\theta$

$z = z$

$x^2 + y^2 = r^2$

Spherical Coordinates:

$x = \rho\sin\phi\cos\theta$

$y = \rho\sin\phi\sin\theta$

$z = \rho\cos\phi$

$x^2 + y^2 + z^2 = \rho^2$

Problem 17: $x^2 + y^2 = 9$

Since $x^2 + y^2 = r^2$, we can directly substitute to get $r^2 = 9$, which means $r = 3$ (since $r \ge 0$).

Problem 18: $x^2 - y^2 = 25$

Substituting $x = r\cos\theta$ and $y = r\sin\theta$, we get $(r\cos\theta)^2 - (r\sin\theta)^2 = 25$.

This simplifies to $r^2\cos^2\theta - r^2\sin^2\theta = 25$, or $r^2(\cos^2\theta - \sin^2\theta) = 25$.

Using the trigonometric identity $\cos(2\theta) = \cos^2\theta - \sin^2\theta$, we have $r^2\cos(2\theta) = 25$.

Problem 19: $x^2 + y^2 + 4z^2 = 10$

Substituting $x^2 + y^2 = r^2$, we get $r^2 + 4z^2 = 10$.

Problem 20: $x^2 + y^2 + 4z^2 = 10$

We have $x^2 + y^2 = (\rho\sin\phi\cos\theta)^2 + (\rho\sin\phi\sin\theta)^2 = \rho^2\sin^2\phi(\cos^2\theta + \sin^2\theta) = \rho^2\sin^2\phi$.

Also, $z^2 = (\rho\cos\phi)^2 = \rho^2\cos^2\phi$.

Thus, $x^2 + y^2 + 4z^2 = \rho^2\sin^2\phi + 4\rho^2\cos^2\phi = 10$.

$\rho^2(\sin^2\phi + 4\cos^2\phi) = 10$.

Problem 21: $2x^2 + 2y^2 - 4z^2 = 0$

We have $2(x^2 + y^2) - 4z^2 = 0$.

Substituting $x^2 + y^2 = \rho^2\sin^2\phi$ and $z^2 = \rho^2\cos^2\phi$, we get $2\rho^2\sin^2\phi - 4\rho^2\cos^2\phi = 0$.

Dividing by $2\rho^2$, we get $\sin^2\phi - 2\cos^2\phi = 0$, or $\sin^2\phi = 2\cos^2\phi$.

Dividing by $\cos^2\phi$, we get $\tan^2\phi = 2$, so $\tan\phi = \sqrt{2}$, and $\phi = \arctan(\sqrt{2})$.

3. Final Answer

1

7. $r = 3$

1

8. $r^2\cos(2\theta) = 25$

1

9. $r^2 + 4z^2 = 10$

2

0. $\rho^2(\sin^2\phi + 4\cos^2\phi) = 10$

2

1. $\tan^2\phi = 2$ or $\phi = \arctan(\sqrt{2})$