The problem requires converting equations from one coordinate system to another. We will solve problems 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, and 27.

1. Problem Description

The problem requires converting equations from one coordinate system to another. We will solve problems 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, and

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7.

2. Solution Steps

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

In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, and $z = z$. Therefore, $x^2 + y^2 = r^2$. Substituting this into the given equation, we have $r^2 = 9$.

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

In cylindrical coordinates, $x = r\cos\theta$ and $y = r\sin\theta$. Therefore, $x^2 - y^2 = r^2\cos^2\theta - r^2\sin^2\theta = r^2(\cos^2\theta - \sin^2\theta) = r^2\cos(2\theta)$.

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

1

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

Since $x^2 + y^2 = r^2$, the equation becomes $r^2 + 4z^2 = 10$.

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

In spherical coordinates, $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, and $z = \rho\cos\phi$. Also, $x^2 + y^2 + z^2 = \rho^2$.

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

Substituting these into the given equation, we have $\rho^2\sin^2\phi + 4\rho^2\cos^2\phi = 10$, or $\rho^2(\sin^2\phi + 4\cos^2\phi) = 10$.

Also, $\rho^2(\sin^2\phi + \cos^2\phi + 3\cos^2\phi) = \rho^2(1 + 3\cos^2\phi) = 10$.

2

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

$2(x^2 + y^2) - 4z^2 = 0$, so $x^2 + y^2 = 2z^2$.

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

Thus, $\sin^2\phi = 2\cos^2\phi$, or $\tan^2\phi = 2$. Therefore, $\tan\phi = \pm\sqrt{2}$.

2

2. $x^2 - y^2 - z^2 = 1$ to spherical coordinates.

$x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, and $z = \rho\cos\phi$.

So, $\rho^2\sin^2\phi\cos^2\theta - \rho^2\sin^2\phi\sin^2\theta - \rho^2\cos^2\phi = 1$.

$\rho^2(\sin^2\phi\cos^2\theta - \sin^2\phi\sin^2\theta - \cos^2\phi) = 1$.

$\rho^2(\sin^2\phi(\cos^2\theta - \sin^2\theta) - \cos^2\phi) = 1$.

$\rho^2(\sin^2\phi\cos(2\theta) - \cos^2\phi) = 1$.

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3. $r^2 + 2z^2 = 4$ to spherical coordinates.

$r^2 = x^2 + y^2 = \rho^2\sin^2\phi$, $z = \rho\cos\phi$.

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

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

$\rho^2(\sin^2\phi + \cos^2\phi + \cos^2\phi) = \rho^2(1 + \cos^2\phi) = 4$.

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4. $\rho = 2\cos\phi$ to cylindrical coordinates.

Multiply both sides by $\rho$: $\rho^2 = 2\rho\cos\phi$.

$\rho^2 = x^2 + y^2 + z^2$, and $z = \rho\cos\phi$.

So, $x^2 + y^2 + z^2 = 2z$.

Then $r^2 + z^2 = 2z$, or $r^2 = 2z - z^2$.

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5. $x + y = 4$ to cylindrical coordinates.

$x = r\cos\theta$, $y = r\sin\theta$.

$r\cos\theta + r\sin\theta = 4$.

$r(\cos\theta + \sin\theta) = 4$.

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6. $x + y + z = 1$ to spherical coordinates.

$x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$.

$\rho\sin\phi\cos\theta + \rho\sin\phi\sin\theta + \rho\cos\phi = 1$.

$\rho(\sin\phi\cos\theta + \sin\phi\sin\theta + \cos\phi) = 1$.

$\rho(\sin\phi(\cos\theta + \sin\theta) + \cos\phi) = 1$.

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

$x^2 + y^2 = \rho^2\sin^2\phi$.

$\rho^2\sin^2\phi = 9$.

3. Final Answer

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7. $r^2 = 9$

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

1

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

2

0. $\rho^2(1 + 3\cos^2\phi) = 10$

2

1. $\tan\phi = \pm\sqrt{2}$

2

2. $\rho^2(\sin^2\phi\cos(2\theta) - \cos^2\phi) = 1$

2

3. $\rho^2(1 + \cos^2\phi) = 4$

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

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5. $r(\cos\theta + \sin\theta) = 4$

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6. $\rho(\sin\phi(\cos\theta + \sin\theta) + \cos\phi) = 1$

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7. $\rho^2\sin^2\phi = 9$