{"id":930,"date":"2025-10-14T16:19:54","date_gmt":"2025-10-14T16:19:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=930"},"modified":"2025-11-13T14:55:34","modified_gmt":"2025-11-13T14:55:34","slug":"alternative-coordinate-systems-and-surfaces-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/alternative-coordinate-systems-and-surfaces-learn-it-2\/","title":{"raw":"Alternative Coordinate Systems and Surfaces: Learn It 2","rendered":"Alternative Coordinate Systems and Surfaces: Learn It 2"},"content":{"raw":"

Spherical Coordinates<\/h2>\r\n

We've seen how cylindrical coordinates extend polar coordinates to 3D by adding a [latex]z[\/latex]-coordinate. Spherical coordinates<\/strong> take a different approach: they use one distance and two angles to locate points in space.<\/p>\r\n

Just as cylindrical coordinates are perfect for cylinders, spherical coordinates excel at describing spheres and problems with spherical symmetry\u2014like analyzing planetary atmospheres, electromagnetic fields radiating from a point, or the volume inside a domed stadium.<\/p>\r\n\r\n

\r\n

spherical coordinate system<\/h3>\r\n

In the\u00a0spherical coordinate system<\/span><\/strong>, a point [latex]P[\/latex] in space (Figure 1) is represented by the ordered triple [latex](\\rho,\\theta,\\varphi)[\/latex] where<\/p>\r\n\r\n

    \r\n \t
  • [latex]\\rho[\/latex] (the Greek letter rho) is the distance between [latex]P[\/latex] and the origin [latex](\\rho\\ne0)[\/latex];<\/li>\r\n \t
  • [latex]\\theta[\/latex] is the same angle used to describe the location in cylindrical coordinates;<\/li>\r\n \t
  • [latex]\\varphi[\/latex] (the Greek letter phi) is the angle formed by the positive [latex]z[\/latex]-axis and line segment [latex]\\overline{OP}[\/latex], where [latex]O[\/latex] is the origin and [latex]0\\leq\\varphi\\leq\\pi[\/latex].<\/li>\r\n<\/ul>\r\n<\/section>\r\n\r\n[caption id=\"attachment_5293\" align=\"aligncenter\" width=\"437\"]\"This Figure 1. The relationship among spherical, rectangular, and cylindrical coordinates.[\/caption]\r\n\r\n
    \r\n
      \r\n \t
    • [latex]\\rho[\/latex] tells you how far the point is from the origin<\/li>\r\n \t
    • [latex]\\theta[\/latex] tells you the direction in the [latex]xy[\/latex]-plane (same as cylindrical)<\/li>\r\n \t
    • [latex]\\varphi[\/latex] tells you the angle down from the positive [latex]z[\/latex]-axis<\/li>\r\n<\/ul>\r\n<\/section>
      The angle [latex]\\varphi[\/latex] is measured from the positive [latex]z[\/latex]-axis, not from the [latex]xy[\/latex]-plane. So [latex]\\varphi = 0[\/latex] points straight up along the [latex]z[\/latex]-axis, [latex]\\varphi = \\frac{\\pi}{2}[\/latex] lies in the [latex]xy[\/latex]-plane, and [latex]\\varphi = \\pi[\/latex] points down toward the negative [latex]z[\/latex]-axis.<\/section>\r\n

      Conversion Formulas<\/h3>\r\n

      The conversion formulas come from careful application of trigonometry to the geometric relationships.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5294\" align=\"aligncenter\" width=\"437\"]\"This Figure 2. The equations that convert from one system to another are derived from right-triangle relationships.[\/caption]\r\n\r\n

      \r\n

      converting among Spherical, cylindrical, and rectangular coordinates<\/h3>\r\n

      Spherical to Rectangular<\/strong>:<\/p>\r\n\r\n

        \r\n \t
      • [latex]x = \\rho \\sin \\varphi \\cos \\theta[\/latex]<\/li>\r\n \t
      • [latex]y = \\rho \\sin \\varphi \\sin \\theta[\/latex]<\/li>\r\n \t
      • [latex]z = \\rho \\cos \\varphi[\/latex]<\/li>\r\n<\/ul>\r\n

        Rectangular to Spherical<\/strong>:<\/p>\r\n\r\n

          \r\n \t
        • [latex]\\rho^2 = x^2 + y^2 + z^2[\/latex] (so [latex]\\rho = \\sqrt{x^2 + y^2 + z^2}[\/latex])<\/li>\r\n \t
        • [latex]\\tan \\theta = \\frac{y}{x}[\/latex] (choose quadrant carefully)<\/li>\r\n \t
        • [latex]\\varphi = \\arccos\\left(\\frac{z}{\\sqrt{x^2 + y^2 + z^2}}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n

          Spherical to Cylindrical<\/strong>:<\/p>\r\n\r\n

            \r\n \t
          • [latex]r = \\rho \\sin \\varphi[\/latex]<\/li>\r\n \t
          • [latex]\\theta = \\theta[\/latex]<\/li>\r\n \t
          • [latex]z = \\rho \\cos \\varphi[\/latex]<\/li>\r\n<\/ul>\r\n

            Cylindrical to Spherical<\/strong>:<\/p>\r\n\r\n

              \r\n \t
            • [latex]\\rho = \\sqrt{r^2 + z^2}[\/latex]<\/li>\r\n \t
            • [latex]\\theta = \\theta[\/latex]<\/li>\r\n \t
            • [latex]\\varphi = \\arccos\\left(\\frac{z}{\\sqrt{r^2 + z^2}}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
              \r\n

              Understanding the Geometry<\/strong>:<\/p>\r\n\r\n

                \r\n \t
              • [latex]z = \\rho \\cos \\varphi[\/latex] comes directly from the right triangle with [latex]\\rho[\/latex] as hypotenuse<\/li>\r\n \t
              • [latex]r = \\rho \\sin \\varphi[\/latex] gives the distance from the [latex]z[\/latex]-axis<\/li>\r\n \t
              • Then [latex]x = r \\cos \\theta = \\rho \\sin \\varphi \\cos \\theta[\/latex] and [latex]y = r \\sin \\theta = \\rho \\sin \\varphi \\sin \\theta[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
                Plot the point with spherical coordinates [latex]\\left(8,\\frac{\\pi}3,\\frac{\\pi}6\\right)[\/latex] and express its location in both rectangular and cylindrical coordinates.[reveal-answer q=\"874652471\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"874652471\"]\r\n

                Use the equations in\u00a0Theorem: Converting among Spherical, Cylindrical, and Rectangular Coordinates\u00a0to translate between spherical and cylindrical coordinates (Figure 4):<\/p>\r\n

                [latex]\\begin{aligned}\r\nx&=\\rho\\sin\\varphi\\cos\\theta=8\\sin\\left(\\frac{\\pi}6\\right)\\cos\\left(\\frac{\\pi}3\\right)=8\\left(\\frac12\\right)\\frac12=2 \\\\\r\ny&=\\rho\\sin\\varphi\\sin\\theta=8\\sin\\left(\\frac{\\pi}6\\right)\\sin\\left(\\frac{\\pi}3\\right)=8\\left(\\frac12\\right)\\frac{\\sqrt{3}}2=2\\sqrt3 \\\\\r\nz&=\\rho\\cos\\varphi=8\\cos\\left(\\frac{\\pi}6\\right)=8\\left(\\frac{\\sqrt{3}}2\\right)=4\\sqrt3.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_5304\" align=\"aligncenter\" width=\"676\"]\"This Figure 4. The projection of the point in the [latex]xy[\/latex]-plane is [latex]4[\/latex] units from the origin. The line from the origin to the point\u2019s projection forms an angle of [latex]\\frac{\\pi}{3}[\/latex] with the positive [latex]x[\/latex]-axis. The point lies [latex]4\\sqrt3[\/latex] units above the [latex]xy[\/latex]-plane.[\/caption]\r\n

                The point with spherical coordinates [latex]\\left(8,\\frac{\\pi}3,\\frac{\\pi}6\\right)[\/latex] has rectangular coordinates [latex](2,2\\sqrt3,4\\sqrt3)[\/latex].<\/p>\r\n

                Finding the values in cylindrical coordinates is equally straightforward:<\/p>\r\n

                [latex]\\begin{aligned}\r\nr&=\\rho\\sin\\varphi=8\\sin\\frac{\\pi}6=4 \\\\\r\n\\theta&=\\theta \\\\\r\nz&=\\rho\\cos\\varphi=8\\cos\\frac{\\pi}6=4\\sqrt3\r\n\\end{aligned}[\/latex].<\/p>\r\n

                Thus, cylindrical coordinates for the point are [latex]\\left(4,\\frac{\\pi}3,4\\sqrt3\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

                Convert the rectangular coordinates [latex](-1,1,\\sqrt6)[\/latex] to both spherical and cylindrical coordinates.[reveal-answer q=\"989282734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989282734\"]\r\n

                Start by converting from rectangular to spherical coordinates:<\/p>\r\n

                [latex]\\begin{aligned}\r\n\\rho^2=x^2+y^2+z^2=(-1)^2+1^2+\\left(\\sqrt6\\right)^2=8 \\quad \\tan\\theta&=\\frac{1}{-1} \\\\\r\n\\rho&=2\\sqrt2 \\quad \\theta&=\\arctan(-1)=\\frac{3\\pi}4\r\n\\end{aligned}[\/latex]<\/p>\r\n

                Because [latex](x,y)=(-1,1)[\/latex], then the correct choice for [latex]\\theta[\/latex] is\u00a0[latex]\\frac{3\\pi}4[\/latex].<\/p>\r\n

                There are actually two ways to identify [latex]\\varphi[\/latex]. We can use the equation [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)[\/latex]. A more simple approach, however, is to use equation [latex]z=\\rho\\cos\\varphi[\/latex]. We know that [latex]z=\\sqrt6[\/latex] and [latex]\\rho=2\\sqrt2[\/latex], so<\/p>\r\n

                [latex]\\sqrt6=2\\sqrt2\\cos\\varphi\\text{, so}\\cos\\varphi=\\frac{\\sqrt6}{2\\sqrt2}=\\frac{\\sqrt3}2[\/latex]<\/p>\r\n

                and therefore [latex]\\varphi=\\frac{\\pi}6[\/latex]. The spherical coordinates of the point are [latex](2\\sqrt2,\\frac{3\\pi}4,\\frac{\\pi}6)[\/latex].<\/p>\r\n

                To find the cylindrical coordinates for the point, we need only find\u00a0[latex]r[\/latex]:<\/p>\r\n

                [latex]r=\\rho\\sin\\varphi=2\\sqrt2\\sin\\left(\\frac{\\pi}6\\right)=\\sqrt2[\/latex].<\/p>\r\n

                The cylindrical coordinates for the point are [latex](\\sqrt2,\\frac{3\\pi}4,\\sqrt6)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

                Surfaces in Spherical Coordinates<\/h3>\r\n

                Understanding constant coordinate surfaces helps you visualize the spherical coordinate system:<\/p>\r\n\r\n

                  \r\n \t
                • [latex]\\rho = c[\/latex]: Sphere<\/strong> of radius [latex]c[\/latex] centered at the origin<\/li>\r\n \t
                • [latex]\\theta = c[\/latex]: Half-plane<\/strong> starting at the [latex]z[\/latex]-axis, making angle [latex]c[\/latex] with the [latex]x[\/latex]-axis (same as cylindrical)<\/li>\r\n \t
                • [latex]\\varphi = c[\/latex]: Half-cone<\/strong> making angle [latex]c[\/latex] with the positive [latex]z[\/latex]-axis<\/li>\r\n<\/ul>\r\n[caption id=\"attachment_5296\" align=\"aligncenter\" width=\"623\"]\"This Figure 3. In spherical coordinates, surfaces of the form [latex]\\rho=c[\/latex] are spheres of radius [latex]\\rho[\/latex] (a), surfaces of the form [latex]\\theta=c[\/latex] are half-planes at an angle [latex]\\theta[\/latex] from the [latex]x[\/latex]-axis (b), and surfaces of the form [latex]\\varphi=c[\/latex] are half-cones at an angle [latex]\\varphi[\/latex] from the [latex]z[\/latex]-axis (c).[\/caption]
                  Why Half-Cones?<\/strong>\u00a0Since [latex]\\varphi[\/latex] is restricted to [latex][0, \\pi][\/latex], a constant [latex]\\varphi[\/latex] value only generates half of a complete cone. If you want a full cone, you'd need to allow [latex]\\varphi[\/latex] to range over a larger interval.<\/section>
                  \r\n

                  Describe the surfaces with the given spherical equations.<\/p>\r\n\r\n

                    \r\n \t
                  1. [latex]\\theta=\\frac{\\pi}3[\/latex]<\/li>\r\n \t
                  2. [latex]\\varphi=\\frac{5\\pi}6[\/latex]<\/li>\r\n \t
                  3. [latex]\\rho=6[\/latex]<\/li>\r\n \t
                  4. [latex]\\rho=\\sin\\theta\\sin\\varphi[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"746526133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"746526133\"]\r\n
                      \r\n \t
                    1. The variable [latex]\\theta[\/latex] represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates [latex](\\rho,\\frac{\\pi}3,\\varphi)[\/latex] lie on the plane that forms angle [latex]\\theta=\\frac{\\pi}3[\/latex] with the positive [latex]x[\/latex]-axis. Because [latex]\\rho>0[\/latex], the surface described by equation [latex]\\theta=\\frac{\\pi}3[\/latex] is the half-plane shown in\u00a0Figure 6.\r\n<\/span>\r\n
                      \r\n
                      [caption id=\"attachment_5311\" align=\"aligncenter\" width=\"279\"]\"This Figure 6. The surface described by equation [latex]\\theta=\\frac{\\pi}3[\/latex] is a half-plane.[\/caption]<\/div>\r\n<\/div><\/li>\r\n \t
                    2. Equation [latex]\\varphi=\\frac{5\\pi}6[\/latex] describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring [latex]\\frac{5\\pi}6[\/latex] rad with the positive [latex]z[\/latex]-axis. These points form a half-cone (Figure 7). Because there is only one value for [latex]\\varphi[\/latex] that is measured from the positive [latex]z[\/latex]-axis, we do not get the full cone (with two pieces).\r\n<\/span>\r\n
                      \r\n
                      [caption id=\"attachment_5312\" align=\"aligncenter\" width=\"445\"]\"This Figure 7. The equation [latex]\\varphi=\\frac{5\\pi}6[\/latex] describes a cone.[\/caption]<\/div>\r\n<\/div>\r\n\r\n<\/span>To find the equation in rectangular coordinates, use equation [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)[\/latex].\u00a0\r\n<\/span>\r\n

                      [latex]\\begin{aligned}\r\n\\frac{5\\pi}6&=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right) \\\\\r\n\\cos\\frac{5\\pi}6&=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\\r\n-\\frac{\\sqrt3}2&=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\\r\n\\frac34&=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\\r\n\\frac{3x^2}4+\\frac{3y^2}4+\\frac{3z^2}4&=z^2 \\\\\r\n\\frac{3x^2}4+\\frac{3y^2}4-\\frac{z^2}4&=0.\r\n\\end{aligned}[\/latex]<\/p>\r\nThis is the equation of a cone centered on the [latex]z[\/latex]-axis.<\/li>\r\n \t

                    3. Equation [latex]\\rho=6[\/latex] describes the set of all points [latex]6[\/latex] units away from the origin\u2014a sphere with radius [latex]6[\/latex] (Figure 8).\r\n[caption id=\"attachment_5313\" align=\"aligncenter\" width=\"461\"]\"This Figure 8. Equation [latex]\\rho=6[\/latex] describes a sphere with radius [latex]6[\/latex].[\/caption]<\/li>\r\n \t
                    4. To identify this surface, convert the equation from spherical to rectangular coordinates, using equations [latex]y=\\rho\\sin\\varphi\\sin\\theta[\/latex] and\u00a0[latex]\\rho^2=x^2+y^2+z^2[\/latex]:\r\n

                      [latex]\\begin{aligned}\r\n\\rho&=\\sin\\theta\\sin\\varphi \\\\\r\n\\rho^2&=\\rho\\sin\\theta\\sin\\varphi&\\text{Multiply both sides of the equation by }\\varphi \\\\\r\nx^2+y^2+z^2&=y&\\text{Substitute rectangular variables using the equations above.} \\\\\r\nx^2+y^2-y+z^2&=0&\\text{Subtract }y\\text{ from both sides of the equation.} \\\\\r\nx^2+y^2-y+\\frac14+z^2&=\\frac14&\\text{Complete the square.} \\\\\r\nx^2+\\left(y-\\frac12\\right)^2+z^2&=\\frac14&\\text{Rewrite the middle terms as a perfect square.}\r\n\\end{aligned}[\/latex]<\/p>\r\nThe equation describes a sphere centered at point [latex] (0,\\frac{1}{2},0)[\/latex] with radius [latex]\\frac{1}{2}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

                      Why Use Spherical Coordinates?<\/h3>\r\nSpherical coordinates simplify problems with spherical symmetry. For example, a sphere with equation [latex]x^2 + y^2 + z^2 = c^2[\/latex] in rectangular coordinates becomes simply [latex]\\rho = c[\/latex] in spherical coordinates.\r\n\r\nPhysics problems involving point sources (like gravitational or electric fields) often have natural spherical symmetry.\r\n

                      In geography, latitude and longitude are used to describe locations on Earth\u2019s surface, as shown in\u00a0Figure 9. Although the shape of Earth is not a perfect sphere, we use spherical coordinates to communicate the locations of points on Earth. Let\u2019s assume Earth has the shape of a sphere with radius 4000 mi. We express angle measures in degrees rather than radians because latitude and longitude are measured in degrees.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5317\" align=\"aligncenter\" width=\"483\"]\"This Figure 9. In the latitude\u2013longitude system, angles describe the location of a point on Earth relative to the equator and the prime meridian.[\/caption]\r\n\r\nLet the center of Earth be the center of the sphere, with the ray from the center through the North Pole representing the positive [latex]z[\/latex]-axis. The prime meridian represents the trace of the surface as it intersects the [latex]xz[\/latex]-plane. The equator is the trace of the sphere intersecting the [latex]xy[\/latex]-plane.\r\n\r\n

                      The latitude of Columbus, Ohio, is [latex]40^{\\small\\circ}[\/latex] N and the longitude is [latex]83^{\\small\\circ}[\/latex] W, which means that Columbus is [latex]40^{\\small\\circ}[\/latex] north of the equator. Imagine a ray from the center of Earth through Columbus and a ray from the center of Earth through the equator directly south of Columbus. The measure of the angle formed by the rays is [latex]40^{\\small\\circ}[\/latex]. In the same way, measuring from the prime meridian, Columbus lies [latex]83^{\\small\\circ}[\/latex] to the west. Express the location of Columbus in spherical coordinates.[reveal-answer q=\"233355186\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233355186\"]The radius of Earth is [latex]4,000[\/latex] mi, so [latex]\\rho=4,000[\/latex]. The intersection of the prime meridian and the equator lies on the positive [latex]x[\/latex]-axis. Movement to the west is then described with negative angle measures, which shows that [latex]\\theta=-83^{\\small\\circ}[\/latex], Because Columbus lies [latex]40^{\\small\\circ}[\/latex] north of the equator, it lies [latex]50^{\\small\\circ}[\/latex] south of the North Pole, so [latex]\\varphi-50^{\\small\\circ}[\/latex]. In spherical coordinates, Columbus lies at point [latex](4,000,-83^{\\small\\circ},50^{\\small\\circ})[\/latex].[\/hidden-answer]<\/section>Cylindrical and spherical coordinates give us the flexibility to select a coordinate system appropriate to the problem at hand. The key is choosing the coordinate system that matches your problem's natural symmetry\u2014this can transform complex calculations into simple ones. In the following example, we examine several different problems and discuss how to select the best coordinate system for each one.\r\n\r\n
                      \r\n

                      In each of the following situations, we determine which coordinate system is most appropriate and describe how we would orient the coordinate axes. There could be more than one right answer for how the axes should be oriented, but we select an orientation that makes sense in the context of the problem.\u00a0Note<\/em>: There is not enough information to set up or solve these problems; we simply select the coordinate system (Figure 10).<\/p>\r\n\r\n

                        \r\n \t
                      1. Find the center of gravity of a bowling ball.<\/li>\r\n \t
                      2. Determine the velocity of a submarine subjected to an ocean current.<\/li>\r\n \t
                      3. Calculate the pressure in a conical water tank.<\/li>\r\n \t
                      4. Find the volume of oil flowing through a pipeline.<\/li>\r\n \t
                      5. Determine the amount of leather required to make a football.<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_5319\" align=\"aligncenter\" width=\"731\"]\"This Figure 10. (credit: (a) modification of work by scl hua, Wikimedia, (b) modification of work by DVIDSHUB, Flickr, (c) modification of work by Michael Malak, Wikimedia, (d) modification of work by Sean Mack, Wikimedia, (e) modification of work by Elvert Barnes, Flickr)[\/caption]\r\n\r\n[reveal-answer q=\"757372734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"757372734\"]\r\n
                          \r\n \t
                        1. Clearly, a bowling ball is a sphere, so spherical coordinates would probably work best here. The origin should be located at the physical center of the ball. There is no obvious choice for how the [latex]x[\/latex]-, [latex]y[\/latex]- and [latex]z[\/latex]-axes should be oriented. Bowling balls normally have a weight block in the center. One possible choice is to align the [latex]z[\/latex]-axis with the axis of symmetry of the weight block.<\/li>\r\n \t
                        2. A submarine generally moves in a straight line. There is no rotational or spherical symmetry that applies in this situation, so rectangular coordinates are a good choice. The [latex]z[\/latex]-axis should probably point upward. The [latex]x[\/latex]- and [latex]y[\/latex]-axes could be aligned to point east and north, respectively. The origin should be some convenient physical location, such as the starting position of the submarine or the location of a particular port.<\/li>\r\n \t
                        3. A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation [latex]z=kr[\/latex], where [latex]k[\/latex] is a constant. In spherical coordinates, we have seen that surfaces of the form [latex]\\varphi=c[\/latex] are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form [latex]z^2=\\frac{x^2}{a^2}+\\frac{y^2}{b^2}[\/latex]. In this case, we could choose any of the three. 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 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. Based on this reasoning, cylindrical coordinates might be the best choice. Choose the [latex]z[\/latex]-axis to align with the axis of the cone. The orientation of the other two axes is arbitrary. The origin should be the bottom point of the cone.<\/li>\r\n \t
                        4. A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. In this case, however, we would likely choose to orient our [latex]z[\/latex]-axis with the center axis of the pipeline. The [latex]x[\/latex]-axis could be chosen to point straight downward or to some other logical direction. The origin should be chosen based on the problem statement. Note that this puts the [latex]z[\/latex]-axis in a horizontal orientation, which is a little different from what we usually do. It may make sense to choose an unusual orientation for the axes if it makes sense for the problem.<\/li>\r\n \t
                        5. A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. The [latex]z[\/latex]-axis should align with the axis of the ball. The origin could be the center of the ball or perhaps one of the ends. The position of the [latex]x[\/latex]-axis is arbitrary.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
                          Watch the following video to see the worked solution to above example.