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Triple Integrals for Volumes of Classical Shapes: Boxes, Cylinders, Spheres, and Cones, Slides of Advanced Calculus

Worked-out examples of finding the volumes of classic shapes (boxes, cylinders, spheres, and cones) using triple integrals. The methods of cylindrical and spherical coordinates are illustrated. This resource aims to help readers better understand how to set up and solve triple integrals.

What you will learn

  • How do you find the volume of a rectangular box using triple integrals?
  • How do you find the volume of a sphere using triple integrals in Cartesian, cylindrical, and spherical coordinates?

Typology: Slides

2021/2022

Uploaded on 09/12/2022

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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 find 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 find 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 region Eis given by ZZZ
E
1dV .
A Rectangular Box
A rectangular box can be described by the set of inequalities axb,cyd,pzq. So that
the volume comes out to be length times width times height as expected:
ZZZ
E
1dV =Zb
aZd
cZq
p
1dzdydx = (ba)(dc)(qp).
A Circular Cylinder
The equation for the outer edge of a circular cylinder of radius ais given by x2+y2=a2. If we want
to consider the volume inside such a cylinder with height h, then we are considering the region where
x2+y2a2and 0 zh(in other words between the planes z= 0 and z=h). We already
have bounds on z, so let’s use that as the innermost integral. Now we need bounds for the circular
x2+y2a2in the xy-plane. We can do that in a few different ways:
1. In Cartesian Coordinates:
The solid can be described by the inequalities axa,a2x2ya2x2, 0 zh.
So we find the volume is:
ZZZ
E
1dV =Za
aZa2
x2
a2
x2Zh
0
1dzdydx =Za
a
2ha2x2dx = 2h1
2πa2=πa2h.
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 0 θ2π, 0 ra, 0 zh. So we find the volume is:
ZZZ
E
1dV =Z2π
0Za
0Zh
0
r dzdrdθ =Z2π
0
Za
0
r dr Zh
0
dz = 2π1
2a2h=πa2h.
Either way, we see that we get the expected volume formula.
pf3

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Download Triple Integrals for Volumes of Classical Shapes: Boxes, Cylinders, Spheres, and Cones and more Slides Advanced Calculus in PDF only on Docsity!

Triple Integrals for Volumes of Some Classic Shapes In the following pages, I give some worked out examples where triple integrals are used to find 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 find 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 region E is given by

E

1 dV.

A Rectangular Box A rectangular box can be described by the set of inequalities a ≤ x ≤ b, c ≤ y ≤ d, p ≤ z ≤ q. So that the volume comes out to be length times width times height as expected: ∫ ∫ ∫

E

1 dV =

∫ (^) b

a

∫ (^) d

c

∫ (^) q

p

1 dzdydx = (b − a)(d − c)(q − p).

A Circular Cylinder The equation for the outer edge of a circular cylinder of radius a is given by x^2 + y^2 = a^2. If we want to consider the volume inside such a cylinder with height h, then we are considering the region where x^2 + y^2 ≤ a^2 and 0 ≤ z ≤ h (in other words between the planes z = 0 and z = h). We already have bounds on z, so let’s use that as the innermost integral. Now we need bounds for the circular x^2 + y^2 ≤ a^2 in the xy-plane. We can do that in a few different ways:

  1. In Cartesian Coordinates: The solid can be described by the inequalities −a ≤ x ≤ a, −

a^2 − x^2 ≤ y ≤

a^2 − x^2 , 0 ≤ z ≤ h. So we find the volume is: ∫ ∫ ∫

E

1 dV =

∫ (^) a

−a

∫ √a (^2) −x 2

−√a^2 −x^2

∫ (^) h

0

1 dzdydx =

∫ (^) a

−a

2 h

a^2 − x^2 dx = 2h

πa^2 = πa^2 h.

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.

  1. In Cylindrical Coordinates: A circular cylinder is perfect for cylindrical coordinates! The region x^2 + y^2 ≤ a^2 is very easily described, so that all together the solid can be described by the inequalities 0 ≤ θ ≤ 2 π, 0 ≤ r ≤ a, 0 ≤ z ≤ h. So we find the volume is: ∫ ∫ ∫

E

1 dV =

∫ (^2) π

0

∫ (^) a

0

∫ (^) h

0

r dzdrdθ =

∫ (^2) π

0

∫ (^) a

0

r dr

∫ (^) h

0

dz = 2π

a^2 h = πa^2 h.

Either way, we see that we get the expected volume formula.

A Sphere The equation for the outer edge of a sphere of radius a is given by x^2 + y^2 + z^2 = a^2. If we want to consider the volume inside, then we are considering the regions x^2 + y^2 + z^2 ≤ a^2. We will set up the inequalities in three ways.

  1. In Cartesian Coordinates: Solving for z gives −

a^2 − x^2 − y^2 ≤ z ≤

a^2 − x^2 − y^2. Then the projection of the sphere onto the xy-plane (i.e. the equation you get when you have z = 0 in the sphere equation) is just the circle x^2 + y^2 = a^2. Now we must describe this with inequalities. All together, the solid can be described by the inequalities −a ≤ x ≤ a, −

a^2 − x^2 ≤ y ≤

a^2 − x^2 , −

a^2 − x^2 − y^2 ≤ z ≤

a^2 − x^2 − y^2. So we can find the volume: ∫ ∫ ∫

E

1 dV =

∫ (^) a

−a

∫ √a (^2) −x 2

−√a^2 −x^2

∫ √a (^2) −x (^2) −y 2

a^2 −x^2 −y^2

1 dzdydx =

∫ (^) a

−a

∫ √a (^2) −x 2

−√a^2 −x^2

a^2 − x^2 − y^2 dydx

∫ (^) a

−a

π(a^2 − x^2 ) dx = π(2a^3 −

a^3 ) =

πa^3.

Note: Same note as I made for the circular cylinder concerning skipped steps in the integration.

  1. In Cylindrical Coordinates: The bound on z would still be the same, but we would use polar for x and y. All together, the solid can be described by 0 ≤ θ ≤ 2 π, 0 ≤ r ≤ a, −

√ a^2 −^ r^2 ≤^ z^ ≤ a^2 − r^2. And we get a volume of: ∫ ∫ ∫

E

1 dV =

∫ (^2) π

0

∫ (^) a

0

∫ √a (^2) −r 2

−√a^2 −r^2

r dzdrdθ = 2π

∫ (^) a

0

2 r

a^2 − r^2 dr

= 2π

∫ (^) a 2

0

u du = 2π

a^3 =

πa^3

  1. In Spherical Coordinates: In spherical coordinates, the sphere is all points where 0 ≤ φ ≤ π (the angle measured down from the positive z axis ranges), 0 ≤ θ ≤ 2 π (just like in polar coordinates), and 0 ≤ ρ ≤ a. And we get a volume of: ∫ ∫ ∫

E

1 dV =

∫ (^) π

0

∫ (^2) π

0

∫ (^) a

0

ρ^2 sin(φ)dρdθdφ =

∫ (^) π

0

sin(φ) dφ

∫ (^2) π

0

∫ (^) a

0

ρ^2 dρ = (2)(2π)

a^3

πa^3

In all three cases, we see that we get the expected volume formula.