Triple Integral

Example from class

Express ∫∫(∫_D) dV as six different iterated integrals where D is the solid region bounded by z=0,z=y, and (x^2)=1-y.

Note: The graphics in what follows are meant to give you a clearer understanding of how the region in (ℝ^3) are described. It is not advisible nor helpful to spend alot of time creating diagrams like I have done here using Mathematica, except possibly the sketch of the region itself. Drawing the regions by hand and getting projections in different planes by hand is (typically) the best way to proceed.

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. And since this is the first problem we've done like this, maybe the technology can help us understand the process better.

region=RegionPlot3D[0≤z≤y &&(x^2)≤1-y,{x,-1,1},{y,0,1},{z,0,1},PlotPoints->40,AxesLabel->{"x","y","z"},Mesh->False,PlotStyle->{Opacity[0.75]}]

Now that we have some understanding of the region D, we can begin to describe it in different ways. Six different ways, in fact.

Description 1: Project onto the x y - plane

This is bit "quick-and-dirty", but we can project onto the x y-plane simply be adding the condition that z=0. That means we will have  (x^2)=1-y. I'm doing this mostly to help you see the projection.

projxy=RegionPlot3D[z==0&&(x^2)≤1-y,{x,-1,1},{y,0,1},{z,0,1},PlotPoints->40,AxesLabel->{"x","y","z"},Mesh->False,PlotStyle->{Red}]

Here is a picture with the region and the projection on the x y - plane. You might want to rotate the image so the region in the x y-plane is easier to see.

Show[region,projxy]

Now, we can describe the region, and we have two choices.

Choice 1

This is a less detailed picture than Figure 15.28 (for a different region) in your text. But the idea is the same--we need to figure out limits for x, y, and z that describe the region. I've chosen two points xp and yp just to create a meaningful sketch. If I was doing this by hand I wouldn't need to define the points.

a=-1;b=1;
xp=0.2;
yp=0.3;
traverse=Show[
{Graphics3D[{Thickness[0.02],Blue,Line[{{a,0,0},{b,0,0}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{xp,0,0},{xp,1-(xp^2),0}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{xp,yp,0},{xp,yp,yp}}]}]
}
];

Show[region,traverse]

This leads to the following description of the region:

D={(x,y,z)|-1≤x≤1,0≤y≤1-(x^2),0≤z≤y}

(∫_(-1)^1)(∫_0^(1-(x^2)))(∫_0^y)1\[DifferentialD]z\[DifferentialD]y\[DifferentialD]x

Choice 2

Here's another description of the region:

xp=0.2;
yp=0.4;
traverse=Show[
{Graphics3D[{Thickness[0.02],Blue,Line[{{-1,0,0},{-1,1,0}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{-(Sqrt[1-yp]),yp,0},{(Sqrt[1-yp]),yp,0}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{xp,yp,0},{xp,yp,yp}}]}]
}
];

Show[region,traverse]

This leads to the following description of the region:

D={(x,y,z)|0≤y≤1,-(Sqrt[1-y])≤x≤(Sqrt[1-y]),0≤z≤y}

(∫_0^1)(∫_(-(Sqrt[1-y]))^(Sqrt[1-y]))(∫_0^y)1\[DifferentialD]z\[DifferentialD]x\[DifferentialD]y

Description 2: Project onto the x z - plane

This is bit "quick-and-dirty", but we can project onto the x z-plane simply be adding the condition that y=0.

projxz=RegionPlot3D[y==0&&(x^2)≤1-z,{x,-1,1},{y,0,1},{z,0,1},PlotPoints->40,AxesLabel->{"x","y","z"},Mesh->False,PlotStyle->{Red}]

Here is a picture with the region and the projection on the x y - plane.

Show[region,projxz]

Now, we can describe the region, and we have two choices.

Choice 1

I've chosen to points xp and yp just to create a meaningful sketch. If I was doing this by hand I wouldn't need to define the points.

a=-1;b=1;
xp=0.2;
zp=0.3;
traverse=Show[
{Graphics3D[{Thickness[0.02],Blue,Line[{{a,0,0},{b,0,0}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{xp,0,0},{xp,0,1-(xp^2)}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{xp,zp,zp},{xp,1-(xp^2),zp}}]}]
}
];

Show[region,traverse,projxz]

This leads to the following description of the region:

D={(x,y,z)|-1≤x≤1,0≤z≤1-(x^2),z≤y≤1-(x^2)}

(∫_(-1)^1)(∫_0^(1-(x^2)))(∫_z^(1-(x^2)))1\[DifferentialD]y\[DifferentialD]z\[DifferentialD]x

Choice 2

Here's another description of the region:

xp=0.2;
zp=0.55;
traverse=Show[
{Graphics3D[{Thickness[0.02],Blue,Line[{{-1,0,0},{-1,0,1}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{-(Sqrt[1-zp]),0,zp},{(Sqrt[1-zp]),0,zp}}]}],
Graphics3D[{Thickness[0.02],Blue,Line[{{xp,zp,zp},{xp,1-(xp^2),zp}}]}]
}
];

Show[region,traverse,projxz]

This leads to the following description of the region:

D={(x,y,z)|0≤z≤1,-(Sqrt[1-z])≤x≤(Sqrt[1-z]),z≤y≤1-(x^2)}

(∫_0^1)(∫_(-(Sqrt[1-z]))^(Sqrt[1-z]))(∫_z^(1-(x^2)))1\[DifferentialD]y\[DifferentialD]x\[DifferentialD]z

Description 3: Project onto the y z - plane

I won't draw the sketches here. You should try to draw some sketches based on the shape of the region

Show[region]

Choice 1

D={(x,y,z)|0≤y≤1,0≤z≤y,-(Sqrt[1-y])≤({{x, ≤(Sqrt[1-y])}})}

(∫_0^1)(∫_0^y)(∫_(-(Sqrt[1-y]))^(Sqrt[1-y]))1\[DifferentialD]x\[DifferentialD]z\[DifferentialD]y

Choice 2

D={(x,y,z)|0≤z≤1,z≤y≤1,-(Sqrt[1-y])≤({{x, ≤(Sqrt[1-y])}})}

(∫_0^1)(∫_z^1)(∫_(-(Sqrt[1-y]))^(Sqrt[1-y]))1\[DifferentialD]x\[DifferentialD]y\[DifferentialD]z

Example

The joint density function for random variables X,Y,Z is given by

f(x,y,z)=(Piecewise[{{(1/8)x y z  , 0≤x≤2, 0≤y≤2,0≤z≤2}, {0, otherwise}}])

Find P(x+Y+Z≤1). The probability is given by the integral of f(x,y,z) over the region described by the constraints.

Barry's Solution

First, let's sketch the region. We should be able to do this by hand as well.

region=RegionPlot3D[0≤x≤2&&0≤y≤2&&0≤z≤2&&x+y+z≤1,{x,0,1},{y,0,1},{z,0,1},PlotPoints->40,AxesLabel->{"x","y","z"},Mesh->False,PlotStyle->{Opacity[0.75]}]

Now that we have some understanding of the region D, we can begin to describe it. Here's one way of describing it:

D={(x,y,z)|0≤x≤1,0≤y≤1-x,0≤z≤1-x-y}

(∫_0^1)(∫_0^(1-x))(∫_0^(1-x-y))(1/8)x y z\[DifferentialD]z\[DifferentialD]y\[DifferentialD]x