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15.7: Triple Integrals in Cylindrical and Spherical Coordinates

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Sometimes, you may end up having to calculate the volume of shapes that have cylindrical, conical, or spherical shapes and rather than evaluating such triple integrals in Cartesian coordinates, you can simplify the integrals by transforming the coordinates to cylindrical or spherical coordinates. For this topic, we will learn how to do such transformations then evaluate the triple integrals.

Introduction

As you learned in Triple Integrals in Rectangular Coordinates, triple integrals have three components, traditionally called x, y, and z. When transforming from Cartesian coordinates to cylindrical or spherical or vice versa, you must convert each component to their corresponding component in the other coordinate system.

There are three coordinate systems that we will be considering. The first is the traditional x, y, and z system also know as the Cartesian coordinate the system; the other two are explored below.

Converting to Cylindrical Coordinates

The second set of coordinates is known as cylindrical coordinates. Working in cylindrical coordinates is essentialy the same as working in polar coordinates in two dimensions except we must account for the z-component of the system. When transforming from Cartesian to cylindircal, x and y become their polar counterparts. Recall that x=rcosθ, y=rsinθ, r2=x2+y2, and tanθ=yx. Now, the conversion for z is simply z=z. r and θ create a plane parallel to the xy-plane, and adding the z component simply gives the plane a "height".

Now, say we have the equation r=1 for 0θ<2π. In two dimensions, that would simply give us a circle centered at (0,0) with a radius of 1. By adding the z-axis, the circle has a height of z, which gives it the shape of a cylinder, hence the name cylindrical coordinates.

As seen in Double Integrals in Polar Form, when converting a double integral from Cartesian to polar coordinates, the dA term, dxdy in Cartesian gets converted to its polar equivilent.

Df(x,y)dxdxyDf(rcosθ,rsinθ)rdrdθ

The same conversion happens with triple integrals from Cartesian to cylindrical for the dV term except you must account for the z-axis with a dz term.

Df(x,y,z)dxdydzDf(rcosθ,rsinθ,z)rdrdθdz

Example 15.7.1: Using Cylindrical Coordinates

Convert this triple integral into cylindrical coordinates and evaluate

111x20y0x2dzdydx

Solution

There are three steps that must be done in order to properly convert a triple integral into cylindrical coordinates.

First, we must convert the bounds from Cartesian to cylindrical. By looking at the order of integration, we know that the bounds really look like

x=1x=1y=1x2y=0z=yz=0

Using the Cartesian to cylindrical conversions, we see that the new bounds are

x=1x=1y=1x2y=0z=yz=0θ=πθ=0r=1r=0z=rsinθz=0

Next, we convert the integrand to its cylindrical equivalent

x2r2cosθ

Thirdly, we convert the differentials at the end of the integral to their cylindrical equivalent being careful to denote the correct order of integration

dzdydxrdzdrdθ

Finally, we put it all together, and we have our newly cylindrically-converted integral

π010rsinθ0r2cos2θrdzdrdθ

Now, we actually evaluate the integral

π010rsinθ0r3cos2θdzdrdθ=π010[r3cos2θz]z=rsinθz=0drdθ=π010r4cos2θsinθdrdθ=π0[r55cos2θsinθ]r=1r=0dθ=15π0cos2θsinθdθ

Using u-substitution, we find that the integrand cos2θsinθ integrates to

15[13cos3θ]θ=πθ=0

Which evaluates to

15[13cos3(π)+13cos3(0)]=215

Converting to Spherical Coordinates

Figure 1 shows a visual representation of spherical coordinates. We define ρ as the distance from the origin to point P. The point P and the origin create a line segment we will call ¯OP, O being the origin. θ is the angle in the x-y plane from the projection of ¯OP, which is shown as ¯OQ. ϕ is the angle between the z-axis and ¯OP.

spherical_coordinates.png
Figure 15.7.1: Spherical coordinates (Duane Q. Nykamp) / CC BY-NC-SA 3.0

The conversions of Cartesian to Spherical are as follows

Just as r=x2+y2, ρ=x2+y2+z2 and like with cylindrical, θ=tan1(yx)

spherical_coordinates_cartesian.png
Figure 15.7.2: Spherical coordinates (Duane Q. Nykamp) / CC BY-NC-SA 3.0

As you can see from Figure 2, r=ρsinϕ, and using this and other trigonometric relations visible here, we can find conversions for x, y, and z.

x and y look like their cylindrical counterparts; however r is replaced with ρsinϕ. So x=ρsinϕcosθ and y=ρsinϕsinθ. Also, from the diagrams, we see that z=ρcosϕ.

As for the dV term of a triple integral, when converted to spherical coordinates, it becomes dV=ρ2sinϕdρdϕdθ.

Example 15.7.2: Using Spherical Coordinates

We are going to find the volume between the sphere ρ=cosϕ and the hemisphere ρ=6. A diagram of the shapes is on the right.spherical_examples.jpg

Solution

First we must set up an integral to calculate the volume:

V=θ1θ0ϕ1ϕ0ρ1ρ0dV

Now we replace the dV term and fill in the bounds of integration:

V=θ1=2πθ0=0ϕ1=π2ϕ0=0ρ1=6ρ0=cosϕρ2sinϕdρdϕdθ

From there we evaluate the integral:

V=132π0π20(216cos3ϕ)sinϕdϕdθ=132π0[216cosϕ+cos4ϕ4]π20dθ=132π0(21614)dθ8634(2π)863π2

Example 15.7.3

Michael wants to eat a bowl of Fruity Hoops Cereal. However, he needs to go to the store and get milk for his cereal, and he is unsure of how much milk to buy. He needs your help deciding the appropriate amount to purchase. The volume of his cereal bowl can be represented by the region bounded below by ρ=4cosϕ and bounded above by z=4. Using this information, find how much milk Michael will need to fill his cereal bowl. The units are in ounces.

First off, we want to draw a diagram, representing the situation, in order to assist us with choosing our bounds of integration.

Contributors and Attributions

  • Paul Salessi (UCD)

15.7: Triple Integrals in Cylindrical and Spherical Coordinates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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