Limits in Spherical Coordinates Definition of spherical coordinates ρ = distance to origin, ρ ≥ 0 φ = angle to z-axis, 0 ≤ φ ≤ π θ = usual θ = angle of projection to xy-plane with x-axis, 0 ≤ θ ≤ 2π Easy trigonometry gives: z = ρ cos φ x = ρ sin φ cos θ y = ρ sin φ sin θ. The equations for x and y are most easily deduced by noticing that for r from polar coordinates we have r = ρ sin φ. This implies x = r cos θ = ρ sin φ cos θ, and y = r sin θ = ρ sin φ sin θ. Going the other way: p ρ = z 2 + y 2 + z 2 φ = cos−1 (z/ρ) θ = tan−1 (y/x). Example: (x, y, z) = (1, 0, 0) ⇒ ρ = 1, φ = π/2, θ = 0 (x, y, z) = (0, 1, 0) ⇒ ρ = 1, φ = π/2, θ = π/2 (x, y, z) = (0, 0, 1) ⇒ ρ = 1, φ = 0, θ –can be anything The volume element in spherical coordinates dV = ρ2 sin φ dρ dφ dθ. The figure at right shows how we get this. The volume of the curved box is ∆V ≈ ∆ρ · ρ∆φ · ρ sin φ∆θ = ρ2 sin φ ∆ρ ∆φ ∆θ. Finding limits in spherical coordinates We use the same procedure asRfor R Rrectangular and cylindrical coordinates. To calculate the limits for an iterated integral D dρ dφ dθ over a region D in 3-space, we are integrating first with respect to ρ. Therefore we q 1. Hold φ and θ fixed, and let ρ increase. This gives us a ray going out from the origin. 2. Integrate from the ρ-value where the ray enters D to the ρ-value where the ray leaves D. This gives the limits on ρ. 1 e 3. Hold θ fixed and let φ increase. This gives a family of rays, that form a sort of fan. Integrate over those φ-values for which the rays intersect the region D. e 4. Finally, supply limits on θ so as to include all of the fans which intersect the region D. For example, suppose we start with the circle in the yz-plane of radius 1 and center at (1, 0), rotate it about the z-axis, and take D to be that part of the resulting solid lying in the first octant. z q l To determine the limits of integration, when φ and θ are fixed, the corresponding ray enters the region where ρ = 0 and leaves where ρ = 2 sin φ. As φ increases, with θ fixed, it is the rays between φ = 0 and φ = π/2 that intersect D, since we are only considering the portion of the surface lying in the first octant (and thus above the xy-plane). Again, since we only want the part in the first octant, we only use θ values from 0 to π/2. So the iterated integral is Z π/2 Z π/2 Z 2 sin φ dρ dφ dθ. 0 0 2 y q O First of all, we have to determine the equation of the surface formed by the rotated circle. In the yz-plane, the two coordinates ρ and φ are indicated. To see the relation between them when P is on the circle, we see that also angle OAP = φ, since both the angle φ and OAP are complements of the same angle, AOP . From the right triangle, this shows the relation is ρ = 2 sin φ. As the circle is rotated around the z-axis, the relationship stays the same, so ρ = 2 sin φ is the equation of the whole surface. 0 P 1 A