[PDF] [PDF] Classic Volume Examples using triple integrals

[PDF] Classic Volume Examples using triple integrals

A Sphere The equation for the outer edge of a sphere of radius a is given by x2 + y2 + z2 = a2 If we want to consider the volume inside, then we are considering the regions x2 + y2 + z2 ≤ a2 We will set up the inequalities in three ways 2 1 2 π(a2 − x2)dx = π(2a3 − 2 3 a3) = 4 3 πa3

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The volume of the curved box is r dr dO dz Then r goes in the blank space in the triple integral-the stretching factor is J = r There are six orders of integration (we give two): volume =f f r dr dO dz = f r dr dz dO

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The volume of the curved box is r dr dO dz Then r goes in the blank space in the triple integral-the stretching factor is J = r There are six orders of integration (we give two): volume =f f r dr dO dz = f r dr dz dO

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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 of

cylindrical 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 p

1dzdydx= (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 x

2+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 x

2+y2a2in thexy-plane. We can do that in a few dierent ways:

1.In Cartesian Coordinates:

The solid can be described by the inequalitiesaxa,pa

2x2ypa

2x2, 0zh.

So we nd the volume is:

ZZZ E 1dV=Z a aZ pa 2x2 pa 2x2Z h 0

1dzdydx=Z

a a2hpa

2x2dx= 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

x

2+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 E

1dV=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,pa

2x2ypa

2x2, pa

2x2y2zpa

2x2y2. So we can nd the volume:

ZZZ E 1dV=Z a aZ pa 2x2 pa 2x2Z pa 2x2y2 pa

2x2y21dzdydx=Z

a aZ pa 2x2 pa

2x22pa

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,pa

2r2zpa

2r2. And we get a volume of:

ZZZ E 1dV=Z 2 0Z a 0Z pa 2r2 pa

2r2rdzdrd= 2Z

a 0 2rpa 2r2dr = 2Z a2

0pudu= 223

a3=43 a3

3.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 0

2sin()ddd=Z

0 sin()dZ 2 0 dZ a 0

2d= (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 (and

downward), 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.

1.In Cartesian Coordinates: First we haveha

px

2+y2zh(I got the rst bound by solving

forzin the equation for the cone and simplifying). The projection down on thexy-plane would be the intersection ofz=hand the cone, which is the discx2+y2a2. So the solid can be described by the inequalitiesaxa,pa

2x2ypa

2x2,ha

px

2+y2zh. We

nd the volume is: ZZZ E 1dV=Z a aZ pa 2x2 pa 2x2Z h ha px

2+y21dzdydx=Z

a aZ pa 2x2 pa

2x2hha

px

2+y2dydx

Z a aZ pa 2x2 pa

2x2hdydxZ

a aZ pa 2x2 pa 2x2ha px

2+y2dydx=ha223

ha2=13 ha2: Note: Again I skipped steps in the integration (this would be a messy/hard integration problem, Cartesian coordinates give messy integrals when working with spheres and cones).

2.In Cylindrical Coordinates: The solid can be described by 02, 0ra,ha

rzh.

And we get a volume of:

ZZZ E 1dV=Z 2 0Z a 0Z h ha rrdzdrd= 2Z a 0 hrha r2dr= 2(12 ha2h3aa3) =13 ha2:

3.In Spherical Coordinates: In spherical coordinates, we need to nd the angle,, that the cone

makes with the positivez-axis and we need to nd the range on. Viewing the cone from the side, the angleis part of a right triangle with side lengthsaandh. So tan() =ah on the edge of the cone. Thus, the range is 0tan1ah . The range ondepends on. We do know the 0zh. And sincez=cos(), we can say that 0hcos()=hsec().

So all together we have 0tan1ah

, 02, and 0hsec(). And we get a volume of: ZZZ E 1dV=Z tan1(ah 0Z 2 0Z hsec() 0

2sin()ddd= 2Z

tan1(ah 0Z hsec() 0

2sin()dd

= 2Z tan1(ah 013 h3sec3()sin()d=23 h3Z tan1(ah 0 sec2()tan()d=23 h312 ah 2=13 ha2 In all three cases, we see that we get the expected volume formula.quotesdbs_dbs14.pdfusesText_20