# Change from cartesian coordinates to cylindrical and spheric

Algebra

### Change from cartesian coordinates to cylindrical and spheric

Hello, I have 6 equations in Cartesian coordinates a) change to cylindrical coordinates b) change to spherical coordinate
This book show me the answers but i don't find it
If anyone can help me i will appreciate so much!

1) z = 2 a) z = 2 b)ρcos(Φ) = 2

2) z = 5x² + 5y² a) z = 5r² b)5ρ = cos(Φ)cosec²(Φ)

3) x² + y² + z² = 9 a) r² + z² = 9 b)ρ = 3

4) x² + y² + 2z² = 4 a) r² +2z² = 4 b)ρ² (1 + cos²(Φ) = 4

5) x² - y² -2z² = 1 a) 2z² = r²cos(2θ) b)ρ² (sin²(Φ)cos(2θ) -2cos²(Φ) = 1

6) x² + y² = 2x a) r = 2cos(θ) b)ρsin(Φ) = 2cos(θ)
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### Re: Change from cartesian coordinates to cylindrical and sph

romsek

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Joined: Sun Feb 21, 2021 7:57 am
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### Re: Change from cartesian coordinates to cylindrical and sph

Do you know what "cylindrical coordinates" and "spherical coordinates" are?

Cylindrical coordinates have polar coordinates in the xy-plane, so a point (x, y) is given by the distance from the origin to (x, y), so that $$r= \sqrt{x^2+ y^2}$$, and the angle from the positive x-axis to the line from the origin to the line from the origin to (x, y), so $$\theta= arctan(y/x)$$, and the same z-axis as used in Cartesian coordinates. Going the other way, $$x= r cos(\theta)$$, $$y= r sin(\theta)$$, and, of course, z= z.

Spherical cooridnates designate the point (x, y, z) by $$\rho$$ the distance from the origin to (x, y, z) so $$\rho= \sqrt{x^2+ y^2+ z^2}$$, the angle $$\theta$$ from the x-axis to the line from the origin to the point (x, y, 0) in the xy-plane, so $$\theta= arctan(y/x)$$, and $$\phi$$, the angle from the positive z-axis to that line, so $$\phi= arctan(r/z)$$ (r is the same as in cylindrical coordinates. Going the other way, $$x= \rho cos(\theta) sin(\phi)$$, $$y= \rho sin(\theta) sin(\phi)$$ and $$z= \rho cos(\phi)$$.

Now, you try the other problems yourself!
The first question is about the Cartesian equation z= 2. That is the plane consisting of all point (x, y, 2), parallel to the xy-plane and 2 units above it. Since the "z" coordinate is the same for Cartesian and Cylindrical coordinates, the equation for Cylindrical coordinates is the same, z= 2. For spherical coordinates, $$z= \rho cos(\phi)= 2$$.

The second is about the Cartesian equation $$z= 5x^2+ 5y^2$$. That is the same as $$z= 5(x^2+ y^2)= 5r^2$$. In Cylindrical coordinates that is $$z= 5r^2$$. It is particularly simple because, geometrically, the figure is a cylinder! In spherical coordinates $$z= \rho cos(\phi)= 5(\rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 sin^(\theta) sin^2(\phi))= 5\rho^2 sin^2(\phi)(cos^2(\theta)+ sin^2(\theta))= 5\rho^2 sin^2(\phi)$$. $$\rho cos^(\phi)= 5\rho^2 sin^2(\phi)$$.$$\rho= \frac{cos(\phi)}{5 sin^2(\phi)}$$.
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