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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.
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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 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:
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.
E
1 dV =
∫ (^2) π
0
∫ (^) a
0
∫ (^) h
0
r dzdrdθ =
∫ (^2) π
0
dθ
∫ (^) 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.
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.
√ 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
E
1 dV =
∫ (^) π
0
∫ (^2) π
0
∫ (^) a
0
ρ^2 sin(φ)dρdθdφ =
∫ (^) π
0
sin(φ) dφ
∫ (^2) π
0
dθ
∫ (^) a
0
ρ^2 dρ = (2)(2π)
a^3
πa^3
In all three cases, we see that we get the expected volume formula.