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Triple Integrals for Volumes of Some Classic Shapes In the following pages, I give some worked out examples where triple integrals are used to nd some
classic shapes volumes (boxes, cylinders, spheres and cones) For all of these shapes, triple integrals aren't
needed, but I just want to show you how you could use triple integrals to nd them. The methods ofcylindrical and spherical coordinates are also illustrated. I hope this helps you better understand how
to set up a triple integral. Remember that the volume of a solid regionEis given byZZZ E 1dV.A Rectangular Box
A rectangular box can be described by the set of inequalitiesaxb,cyd,pzq. So that the volume comes out to be length times width times height as expected: ZZZ E 1dV=Z b aZ d cZ q p1dzdydx= (ba)(dc)(qp):
A Circular Cylinder
The equation for the outer edge of a circular cylinder of radiusais given byx2+y2=a2. If we want to consider the volume inside such a cylinder with heighth, then we are considering the region where x2+y2a2and 0zh(in other words between the planes
z= 0 andz=h). We already have bounds onz, so let's use that as the innermost integral. Now we need bounds for the circular x2+y2a2in thexy-plane. We can do that in a few dierent ways:
1.In Cartesian Coordinates:
The solid can be described by the inequalitiesaxa,pa2x2ypa
2x2, 0zh.
So we nd the volume is:
ZZZ E 1dV=Z a aZ pa 2x2 pa 2x2Z h 01dzdydx=Z
a a2hpa2x2dx= 2h12
a2=a2h: Note: I skipped some steps in the integration. You would need to see the last integration geomet-rically (that the last integral represents the area of exactly half a circle), or you would have to use
trig substitution.2.In Cylindrical Coordinates: A circular cylinder is perfect for cylindrical coordinates! The region
x2+y2a2is very easily described, so that all together the solid can be described by the
inequalities 02, 0ra, 0zh. So we nd the volume is: ZZZ E1dV=Z2
0Z a 0Z h 0 rdzdrd=Z 2 0 dZ a 0 rdrZ h 0 dz= 212 a2h=a2h: Either way, we see that we get the expected volume formula.A Sphere
The equation for the outer edge of a sphere of radiusais given byx2+y2+z2=a2. If we want to consider the volume inside, then we are considering the regionsx2+y2+z2a2. We will set up the inequalities in three ways.1.In Cartesian Coordinates: Solving forzgivespa
2x2y2zpa
2x2y2. Then the
projection of the sphere onto thexy-plane (i.e. the equation you get when you havez= 0 in the sphere equation) is just the circlex2+y2=a2. Now we must describe this with inequalities. All together, the solid can be described by the inequalitiesaxa,pa2x2ypa
2x2, pa2x2y2zpa
2x2y2. So we can nd the volume:
ZZZ E 1dV=Z a aZ pa 2x2 pa 2x2Z pa 2x2y2 pa2x2y21dzdydx=Z
a aZ pa 2x2 pa2x22pa
2x2y2dydx
Z a a212 (a2x2)dx=(2a323 a3) =43 a3: Note: Same note as I made for the circular cylinder concerning skipped steps in the integration.2.In Cylindrical Coordinates: The bound onzwould still be the same, but we would use polar for
xandy. All together, the solid can be described by 02, 0ra,pa2r2zpa
2r2. And we get a volume of:
ZZZ E 1dV=Z 2 0Z a 0Z pa 2r2 pa2r2rdzdrd= 2Z
a 0 2rpa 2r2dr = 2Z a20pudu= 223
a3=43 a33.In Spherical Coordinates: In spherical coordinates, the sphere is all points where 0(the
angle measured down from the positivezaxis ranges), 02(just like in polar coordinates), and 0a. And we get a volume of: ZZZ E 1dV=Z 0Z 2 0Z a 02sin()ddd=Z
0 sin()dZ 2 0 dZ a 02d= (2)(2)13
a3 =43 a3 In all three cases, we see that we get the expected volume formula.A Cone
The equationa2z2=h2x2+h2y2gives a cone with a point at the origin that opens upward (anddownward), such that if the height isz=hthen radius of the circle at that height isa(you can see this
by pluggin inz=hand simplifying). So let's nd the volume inside this cone which has heighthand radius ofaat that height.