Triple Integrals

1 / 28

# Triple Integrals - PowerPoint PPT Presentation

Triple Integrals. z-Simple, y-simple, z-simple Approach. z-Simple solids (Type 1). Definition: A solid region E is said to be z-Simple if it is bounded by two surfaces z=z 1 (x,y) and z=z 2 (x,y) (z 1 £ z £ z 2 ). Iterated Triple Integrals over z-Simple solid E.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Triple Integrals' - skyler-lane

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### z-Simple, y-simple, z-simple Approach

z-Simple solids (Type 1)
• Definition:
• A solid region E is said to be z-Simple if it is bounded by two surfaces z=z1(x,y) and z=z2(x,y)(z1 £ z£z2)
Iterated Triple Integrals over z-Simple solid E
• When you project a z-Simple solid E onto the xy-plane you obtain a planar region D.
• 1st you integrate wrt z (the simple variable) from z=z1(x,y) to z=z2(x,y).
• You obtain some function of x and y to integrate over the region D in the xy-plane.
• If D is Type I you have y=y1(x) to y=y2(x) and you integrate over y.
• Finally you integrate over the constant limits from x=x1 to x=x2 and this integration is wrt x.
• If D is Type II you have x=x1(y) to x=x2(y).
• Finally you integrate over the constant limits from y=y1 to y=y2 and this integration is wrt y.
z-Simple solid: special caser=1, z1=0, z2>0.

This gives the volume V over the region Dxy in the xy-plane of the surface z=z2(x,y)

y-Simple solids (Type 2)
• Definition:
• A solid region E is said to be y-Simple if it is bounded by two surfaces y=y1(x,z) and y=y2(x,y)(y1 £y£y2)
Example: Paraboloids
• Find the volume of the solid enclosed by the two surfacesy= 0.5(x2+z2) and y=16-x2-z2.
• We need to define a region Dxz in the xz-plane.
Example
• Eliminate y by equating0.5(x2+z2) =16-x2-z2.
• This gives x2+z2 =32/3
It may be helpful to recall the single integral calculus method for finding area between two curves y=y1(x) and y=y2(x) -- just think of z as constant, e.g., on a horizontal trace (say z=0)
• Next we evaluatethe integral -- dy goesfirst since it is y-simple.

Polar coordinates x=r cos q,z=r sin q, were

used so that dAxz can be written as

r dr dq instead ofdx dz.

Compare to a cylinder of radius and height 16 which has

double this volume (anyone know why?) and contains our solid E inside it.

### xy-Simple, yz-simple, xz-simple Approach

Often a solid is simple in more than one variable.An alternate approach is to look for the one variable that it is not simple in, and make that the outer limitof integration. The inner limit is then a double integral.

This approach is also helpful in sketching the solid of integration, because as we will see the outer limit ofintegration corresponds to constant values on whichcontour regions in the simple plane lie

If not z-simple, try:

where Dxy(z) is the trace of the solid (a trace of a solid is a region instead of a curve) in the plane z=constant.

Example (Text, page 892#7)
• Sketch the domain of integration of the triple integralwhereThen evaluate the integral.
Solution:
• Perhaps the easiest way to see the solid E which is our domain of integration is to first consider z fixed.
• Then x varies from 0 to (z + 2) and y varies from 0 to 2z.
• This defines a rectangle in the plane z units above the xy-plane.
• Question: What are its vertices?
• Answer: (0,0,z), (z + 2, 0, z), (0, 2z, z) and (z+2, 2z,z).
Horizontal trace of domain E (z=constant)Rectangle (0,0,z), (z + 2, 0, z), (0, 2z, z), (z+2, 2z,z).
• This rectangle lies on a plane which is located z units above the xy-plane.
• Let’s graph the family of horizontal traces, for several values of z between 0 and 1.
Domain E is not z-simple
• Next, as z increases, the rectangles become larger (and higher)
• If we stack them, one above the other, we get the solid domain of integration
• Let’s assume r(x,y,z)=1, so that the triple integral will give us a volume instead of a mass.
Cross-section Rxy(keeping z constant)
• We integrate r(x,y,z)=1 over Rxy treating z as constant to get the area of Rxy
• Its horizontal traces define regions R=Rxy(z) above the xy-plane. Each of these planar regions should be Type I (or Type II) so that the areas of the cross-sections can be evaluated as double integrals
• Between z and z+dz the total volume is dV=Axydz. Sum from z=0 to 1 to get the total volume: that sum converges to the integral over z.
Challenge
• Rewrite
• as
• or
• using the fact that E is y-simple (i.e., dy on the inside)
Answers: dy dx dz is easy
• but dy dz dx has to be split into 2.
Example: Tetrahedron(x,y, and z-simple :-)
• Evaluate the triple integral
• where E is the solid tetrahedron with vertices (0,0,0), (1,0,0), (0,2,0), and (0,0,3).
Example: Step 1
• Visualize the solid. You need to get equations of the 4 planar sides of the tetrahedron.
• Consider P(0,0,0), Q(1,0,0), R(0,2,0)Convince yourself -- or show usingthat the equation of the plane through these 3 points is z=0.
Consider P(0,0,0), Q(1,0,0), S(0,0,3)
• Similarly the equation of the plane through these 3 points is y=0.
• Consider P(0,0,0), R(0,2,0), S(0,0,3)
• This corresponds the plan x=0.
• Now consider Q(1,0,0), R(0,2,0), S(0,0,3)
Example: Step 2
• The region is described by 0 £ x, 0 £ y, 0 £ z, and 6x+3y+2z £ 6.
• This solid is x-simple, y-simple and z-simple.
• To describe it as z-simple, we let z1=0 and z2=3-3x-1.5y.
• Equating z=z1=z2 we obtain the region Dxy described by 3x+1.5y=3 in the xy-plane.
• Since 0 £ y £ 2-2x for 0 £ x £ 1 ...
Example: Step 3
• We now are in the position to set up the triple integral with limits.