Triple Integrals in Spherical and Cylindrical Coordinates

Example from class

Evaluate (∫_0^3)(∫_0^(Sqrt[9-(y^2)]))(∫_(Sqrt[(x^2)+(y^2)])^(Sqrt[18-(x^2)-(y^2)]))((x^2)+(y^2)+(z^2))dzdxdy by changing to spherical coordinates.

Sketch

First, let's sketch the region. We should be able to do as many of these sketches as possible by hand, but if not we should be able to use technology to get them.

region=RegionPlot3D[(x^2)+(y^2)+(z^2)≤18&&(z^2)≥(x^2)+(y^2),{x,0,3},{y,0,3},{z,0,(Sqrt[18])},PlotPoints->40,AxesLabel->{"x","y","z"},Mesh->False,PlotStyle->{Opacity[0.75]}]

Doing the Integral

Here's the change we worked out in class:

(∫_0^3)(∫_0^(Sqrt[9-(y^2)]))(∫_(Sqrt[(x^2)+(y^2)])^(Sqrt[18-(x^2)-(y^2)]))((x^2)+(y^2)+(z^2))\[DifferentialD]z\[DifferentialD]x\[DifferentialD]y

(∫_0^(Pi/4))(∫_0^(Pi/2))(∫_0^(Sqrt[18]))((ρ^4)Sin[ϕ])\[DifferentialD]ρ\[DifferentialD]θ\[DifferentialD]ϕ

The second integral is much easier to work out than the first.

(∫_0^(Pi/4))(∫_0^(Pi/2))(∫_0^(Sqrt[18]))((ρ^4)Sin[ϕ])dρdθdϕ
=(∫_0^(Pi/4))(∫_0^(Pi/2))((((((Sqrt[18])))^5)/5)Sin[ϕ])dθdϕ
=(((((Sqrt[18])))^5)/5)(∫_0^(Pi/4))(∫_0^(Pi/2))(Sin[ϕ])dθdϕ
=(Pi/2)(((((Sqrt[18])))^5)/5)(∫_0^(Pi/4))(Sin[ϕ])dϕ
=Pi(((((Sqrt[18])))^5)/10)(Subscript[(-Cos[ϕ]), 0]^(Pi/4))
=2Pi(((((Sqrt[18])))^5)/10)(-Cos[Pi/4]+Cos[0])
=2Pi(((((Sqrt[18])))^5)/10)(-(1/(Sqrt[2]))+1) (this answer is fine, but you can simplify if you like to verify Mathematica's answer)
=Pi((972 (Sqrt[2]))/10)(-(1/(Sqrt[2]))+1)
=Pi((486 )/5)(-1+(Sqrt[2]))

Example from class

Evaluate ∫∫(∫_E)((x^2))dV where E is the solid that lies between the cylinder (x^2)+(y^2)=1 and above the plane z=0, and below the cone (z^2)=4(x^2)+4(y^2).

Sketch

First, let's sketch the region. We should be able to do as many of these sketches as possible by hand, but if not we should be able to use technology to get them.

region=RegionPlot3D[(x^2)+(y^2)≤1&&(z^2)≤4(x^2)+4(y^2)&&z≥0,{x,-1,1},{y,-1,1},{z,0,2},PlotPoints->50,AxesLabel->{"x","y","z"},Mesh->False,PlotStyle->{Opacity[0.75]}]

Do you think the fact that the region looks choppy at the top is a problem? Why do you think that is happening?

Doing the Integral

Here's the integral we worked out in class, using cylindrical coordinates:

(∫_0^(2Pi))(∫_0^1)(∫_0^(2r))((r^3)((Cos[θ])^2))\[DifferentialD]z\[DifferentialD]r\[DifferentialD]θ

Here are the details of how the integral is done:

(∫_0^(2Pi))(∫_0^1)(∫_0^(2r))((r^3)((Cos[θ])^2))dzdrdθ
=(∫_0^(2Pi))(∫_0^1)(2(r^4)((Cos[θ])^2))drdθ
=(∫_0^(2Pi))(Subscript[(2((r^5)/5)((Cos[θ])^2)), r=0]^(r=1))dθ
=(2/5)(∫_0^(2Pi))((Cos[θ])^2)dθ
=(2/5)(∫_0^(2Pi))(1/2)(1+Cos[2θ])dθ
=(1/5)(Subscript[(θ+(1/2)Sin[2θ]), 0]^(2Pi))
=(1/5)(2Pi+(1/2)Sin[4Pi]-(1/2)Sin[0])
=((2Pi)/5)

Example from class

Find the volume of the solid that lies above the cone ϕ=π/3 and below the sphere ρ=4 cos ϕ.

Sketch

First, let's sketch the region. We should be able to do as many of these sketches as possible by hand, but if not we should be able to use technology to get them.

Clear[rho,theta,phi]
phi= Pi/3;
surface1=ParametricPlot3D[{rho Sin[phi] Cos[theta],rho Sin[phi] Sin[theta],rho Cos[phi]},{theta,0,2Pi},{rho,0,2}]

Clear[rho,theta,phi]
rho= 4Cos[phi];
surface2=ParametricPlot3D[{rho Sin[phi] Cos[theta],rho Sin[phi] Sin[theta],rho Cos[phi]},{theta,0,2Pi},{phi,0,Pi},Mesh->False,PlotStyle->Opacity[0.25]]

Show[surface2,surface1]

Notice that this would be a fairly tricky region to describe in cartesian coordinates, since the sphere extends out past the cone on the sides.

Doing the Integral

Here's the integral we worked out in class, using spherical coordinates:

Clear[rho,theta,phi,ρ,θ,ϕ]
(∫_0^(Pi/3))(∫_0^(2Pi))(∫_0^(4Cos[ϕ]))(ρ^2)Sin[ϕ]\[DifferentialD]ρ\[DifferentialD]θ\[DifferentialD]ϕ

Here are the details of how the integral is done:

V=(∫_0^(Pi/3))(∫_0^(2Pi))(∫_0^(4Cos[ϕ]))(ρ^2)Sin[ϕ]dρdθdϕ
=(∫_0^(Pi/3))(∫_0^(2Pi))(Subscript[(((ρ^3)/3)Sin[ϕ]), 0]^(4Cos[ϕ]))dθdϕ
=(∫_0^(Pi/3))(∫_0^(2Pi))(((64((Cos[ϕ])^3))/3)Sin[ϕ])dθdϕ
=(64/3)(∫_0^(Pi/3))(∫_0^(2Pi))(((Cos[ϕ])^3)Sin[ϕ])dθdϕ
=((128π)/3)(∫_0^(Pi/3))(((Cos[ϕ])^3)Sin[ϕ])dϕ

Let u=Cos[ϕ], du = -Sin[ϕ] dϕ
when ϕ=0, u=1 and when ϕ=Pi/3, u=(1/2)

=-((128π)/3)(∫_1^(1/2))(u^3)du
=-((128π)/3)(Subscript[(((u^4)/4)), 1]^(1/2))
=-((32π)/3)(Subscript[((u^4)), 1]^(1/2))
=-((32π)/3)(((((1/2)))^4)-1)
=-((32π)/3)((1/16)-1)
=-((32π)/3)(-(15/16))
=2π(5)
=10π