Triple Integrals in Cylindrical or Spherical Coordinates

Transcription

1 Triple Integrals in Clindrical or Spherical Coordinates. Find the volume of the solid ball Solution. Let be the ball. We know b #a of the worksheet Triple Integrals that the volume of is given b the triple integral dv. To compute this, we need to convert the triple integral to an iterated integral. The given ball can be described easil in spherical coordinates b the inequalities ρ, φ π, θ < 2π, so we can rewrite the triple integral dv as an iterated integral in spherical coordinates 2π π ρ 2 sin φ dρ dφ dθ 2π π 2π π 2π 2π 4 π ρ sin φ φ ρ ρ sin φ dφ dθ φπ cos φ dθ 2 dθ dφ dθ 2. Let be the solid inside both the cone and the sphere Write the triple integral dv as an iterated integral in spherical coordinates. Solution. Here is a picture of the solid: We have to write both the integrand and the solid of integration in spherical coordinates. We know that in Cartesian coordinates is the same as ρ cos φ in spherical coordinates, so the function we re integrating is ρ cos φ. The cone is the same as φ π 4 in spherical coordinates. The sphere is ρ in spherical coordinates. So, the solid can be described in spherical coordinates as ρ, φ π 4, θ 2π. This means that the iterated integral is 2π π/4 ρ cos φρ 2 sin φ dρ dφ dθ. Wh? We could first rewrite in clindrical coordinates: it s r. In terms of spherical coordinates, this sas that ρ cos φ ρ sin φ, so cos φ sin φ. That s the same as saing that tan φ, or φ π 4.

2 For the remaining problems, use the coordinate sstem Cartesian, clindrical, or spherical that seems easiest.. Let be the ice cream cone bounded below b and above b Write an iterated integral which gives the volume of. You need not evaluate. Solution. We know b #a of the worksheet Triple Integrals that the volume of is given b the triple integral dv. The solid has a simple description in spherical coordinates, so we will use spherical coordinates to rewrite the triple integral as an iterated integral. The sphere is the same as ρ 2. The cone can be written as φ π 6.2 So, the volume is 2π π/6 2 ρ 2 sin φ dρ dφ dθ. 4. Write an iterated integral which gives the volume of the solid enclosed b ,, and 2. You need not evaluate. Solution. We know b #a of the worksheet Triple Integrals that the volume of is given b the triple integral dv. To compute this, we need to convert the triple integral to an iterated integral. Since the solid is smmetric about the -ais but doesn t seem to have a simple description in terms of spherical coordinates, we ll use clindrical coordinates. Let s think of slicing the solid, using slices parallel to the -plane. This means we ll write the outer integral first. We re slicing [, 2] on the -ais, so our outer integral will be 2 something d. To write the inner double integral, we want to describe each slice and, within a slice, we can think of as being a constant. Each slice is just the disk enclosed b the circle , which is a circle 2 This is true because can be written in clindrical coordinates as r. In terms of spherical coordinates, this sas that ρ cos φ ρ sin φ. That s the same as saing tan φ, or φ π 6. 2

3 of radius : We ll use polar coordinates to write the iterated double integral describing this slice. The circle can be described as θ < 2π and r and remember that we are still thinking of as a constant, so the appropriate integral is 2π r dr dθ. Putting this into our outer integral, we get the iterated integral 2 2π r dr dθ d. Note: For this problem, writing the inner integral first doesn t work as well, at least not if we want to write the integral with d as the inner integral. Wh? Well, if we tr to write the integral with d as the inner integral, we imagine sticking vertical lines through the solid. The problem is that there are different tpes of vertical lines. For instance, along the red line in the picture below, goes from the cone or r to 2 in the solid. But, along the blue line, goes from to 2. So, we d have to write two separate integrals to deal with these two different situations. 5. Let be the solid enclosed b and 9. Rewrite the triple integral iterated integral. You need not evaluate, but can ou guess what the answer is? dv as an Solution describes a paraboloid, so the solid looks like this: Since the solid is smmetric about the -ais, a good guess is that clindrical coordinates will make

4 things easier. In clindrical coordinates, the integrand is equal to r cos θ. Let s think of slicing the solid, which means we ll write our outer integral first. If we slice parallel to the -plane, then we are slicing the interval [, 9] on the -ais, so our outer integral is 9 something d. We use the inner two integrals to describe a tpical slice; within a slice, is constant. Each slice is a disk enclosed b the circle which has radius. We know that we can describe this in polar coordinates as r, θ < 2π. So, the inner two integrals will be Therefore, the given triple integral is equal to the iterated integral 9 2π 9 2π r cos θ r dr dθ d r cos θ 9 2π 9 2π r r /2 cos θ dθ d θ2π /2 sin θ d θ dr dθ d r cos θ r dr dθ. That the answer is should not be surprising because the integrand f,, is anti-smmetric about the plane this is sort of like saing the function is odd: f,, f,,, but the solid is smmetric about the plane. Note: If ou decided to do the inner integral first, ou probabl ended up with d as our inner integral. In this case, a valid iterated integral is 6. The iterated integral in spherical coordinates 2π 9 π π/2 r 2 r cos θ r d dr dθ. π/2 2 solid. Describe the solid its shape and its densit at an point. ρ sin φ dρ dφ dθ computes the mass of a Solution. The shape of the solid is described b the region of integration. We can read this off from the bounds of integration: it is π 2 θ π, φ π 2, ρ 2. We can visualie ρ 2 b imagining a solid ball of radius 2 with a solid ball of radius taken out of the middle. φ π 2 tells us we ll onl have the top half of that, and π 2 θ π tells us that we ll onl be looking at one octant: the one with negative and positive: To figure out the densit, remember that we get mass b integrating the densit. If we call this solid, then the iterated integral in the problem is the same as the triple integral ρ sin 2 φ dv since 4

5 dv is ρ 2 sin φ dρ dφ dθ. So, the densit of the solid at a point ρ, φ, θ is ρ sin 2 φ. 5