The terms involving 1/r are called metric or curvature. For this part, we simplify substitute inand, hence, for the material derivative of the velocity vector in spherical coordinates: Dv Dt Du Dt-uv r tan + uw r i + Dv Dt + u2 r tan + vw r j+ Dw Dt-u2 +v2 r k (4.17) where theicomponent is associated with Du Dt, thej component with Dv Dt and the k component with Dw Dt. 2.7.1 Convert from cylindrical to rectangular coordinates.Spherical Polar Coordinates This integral has three main parts: ZZZ V F(x,y,z)dV (8) 1. I Notice the extra factor 2 sin() on the right-hand side. Remark: I Spherical coordinates are useful when the integration region R is described in a simple way using spherical coordinates. Integral of function f in the region R can be expressed in spherical coordinates as follows, ZZZ R f dv ZZZ R f (,) 2 sin() d d d.

Some surfaces, however, can be difficult to model with equations based on the Cartesian system. 2.7.4 Convert from rectangular to spherical coordinates.The Cartesian coordinate system provides a straightforward way to describe the location of points in space. A spherical coordinate system is useful when there is a point of symmetry that. 1-1-3 Spherical Coordinates.

Similarly, spherical coordinates are useful for dealing with problems involving spheres, such as finding the volume of domed structures. As the name suggests, cylindrical coordinates are useful for dealing with problems involving cylinders, such as calculating the volume of a round water tank or the amount of oil flowing through a pipe. In this section, we look at two different ways of describing the location of points in space, both of them based on extensions of polar coordinates.

X = r cos θ These equations are used to convert from y = r sin θ cylindrical coordinates to rectangular z = z coordinates. And r 2 = x 2 + y 2 These equations are used to convert from tan θ = y x rectangular coordinates to cylindrical z = z coordinates. In this way, cylindrical coordinates provide a natural extension of polar coordinates to three dimensions.X = r cos θ These equations are used to convert from y = r sin θ cylindrical coordinates to rectangular z = z coordinates. Starting with polar coordinates, we can follow this same process to create a new three-dimensional coordinate system, called the cylindrical coordinate system.

Right-triangle relationships tell us that x = r cos θ , x = r cos θ , y = r sin θ , y = r sin θ , and tan θ = y / x. To make this easy to see, consider point P P in the xy-plane with rectangular coordinates ( x , y , 0 ) ( x , y , 0 ) and with cylindrical coordinates ( r , θ , 0 ) , ( r , θ , 0 ) , as shown in the following figure.Figure 2.90 The Pythagorean theorem provides equation r 2 = x 2 + y 2. Notice that these equations are derived from properties of right triangles. Note that if x = 0 , x = 0 , then the value of θ θ is either π 2 , 3 π 2 , π 2 , 3 π 2 , or 0 , 0 , depending on the value of y. However, if we restrict θ θ to values between 0 0 and 2 π , 2 π , then we can find a unique solution based on the quadrant of the xy-plane in which original point ( x , y , z ) ( x , y , z ) is located.

Now, let’s think about surfaces of the form r = c. Therefore, in cylindrical coordinates, surfaces of the form z = c z = c are planes parallel to the xy-plane. When we convert to cylindrical coordinates, the z-coordinate does not change. Planes of these forms are parallel to the yz-plane, the xz-plane, and the xy-plane, respectively. If c c is a constant, then in rectangular coordinates, surfaces of the form x = c , x = c , y = c , y = c , or z = c z = c are all planes.

$dv in spherical coordinates$

Then, looking at the triangle in the xy-plane with r r as its hypotenuse, we have x = r cos θ = ρ sin φ cos θ. Looking at Figure 2.98, it is easy to see that r = ρ sin φ. And ρ = r 2 + z 2 These equations are used to convert from θ = θ cylindrical coordinates to spherical φ = arccos ( z r 2 + z 2 ) coordinates.The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they are straightforward applications of trigonometry.

Also, note that, as before, we must be careful when using the formula tan θ = y x tan θ = y x to choose the correct value of θ. Φ = arccos ( z r 2 + z 2 ). Solving this last equation for φ φ and then substituting ρ = r 2 + z 2 ρ = r 2 + z 2 (from the first equation) yields φ = arccos ( z r 2 + z 2 ). Figure 2.96 also shows that ρ 2 = r 2 + z 2 = x 2 + y 2 + z 2 ρ 2 = r 2 + z 2 = x 2 + y 2 + z 2 and z = ρ cos φ. The derivation of the formula for y y is similar.

One possible choice is to align the z-axis with the axis of symmetry of the weight block. Bowling balls normally have a weight block in the center. There is no obvious choice for how the x-, y- and z-axes should be oriented. The origin should be located at the physical center of the ball.

The origin should be some convenient physical location, such as the starting position of the submarine or the location of a particular port. The x- and y-axes could be aligned to point east and north, respectively. The z-axis should probably point upward. There is no rotational or spherical symmetry that applies in this situation, so rectangular coordinates are a good choice.

Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form z 2 = x 2 a 2 + y 2 b 2. In spherical coordinates, we have seen that surfaces of the form φ = c φ = c are half-cones. In cylindrical coordinates, a cone can be represented by equation z = k r , z = k r , where k k is a constant.

Choose the z-axis to align with the axis of the cone. Based on this reasoning, cylindrical coordinates might be the best choice. In addition, we are talking about a water tank, and the depth of the water might come into play at some point in our calculations, so it might be nice to have a component that represents height and depth directly. However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might want to avoid that choice.

In this case, however, we would likely choose to orient our z-axis with the center axis of the pipeline. A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. The origin should be the bottom point of the cone.

The z-axis should align with the axis of the ball. A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. It may make sense to choose an unusual orientation for the axes if it makes sense for the problem. Note that this puts the z-axis in a horizontal orientation, which is a little different from what we usually do. The origin should be chosen based on the problem statement.

Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. We recommend using aAuthors: Gilbert Strang, Edwin “Jed” Herman© OpenStax. The position of the x-axis is arbitrary.Want to cite, share, or modify this book? This book isCreative Commons Attribution-NonCommercial-ShareAlike License 4.0If you are redistributing all or part of this book in a print format,Then you must include on every physical page the following attribution:If you are redistributing all or part of this book in a digital format,Then you must include on every digital page view the following attribution:Use the information below to generate a citation.

Dv In Spherical Coordinates