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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.

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z simple solids type 1
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
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 case r 1 z 1 0 z 2 0
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
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
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
Example
  • Eliminate y by equating0.5(x2+z2) =16-x2-z2.
  • This gives x2+z2 =32/3
slide10
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.
slide11

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

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
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
Example (Text, page 892#7)
  • Sketch the domain of integration of the triple integralwhereThen evaluate the integral.
slide15
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
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
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 r xy keeping z constant
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
Challenge
  • Rewrite
  • as
  • or
  • using the fact that E is y-simple (i.e., dy on the inside)
slide21
Answers: dy dx dz is easy
  • but dy dz dx has to be split into 2.
example tetrahedron x y and z simple
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
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.
slide24
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
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
Example: Step 3
  • We now are in the position to set up the triple integral with limits.