4.8: volumes of solids of revolution - dartmouth college of a right prism right prism with base...

4.8: Volumes of Solids of Revolution

Mathematics 3Lecture 25

Dartmouth College

March 03, 2010

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Example 1

Before we begin, let’s test our knowledge of integration. Consider theindefinite integral: ∫

x√x+ 1 dx

a.) Evaluate by using integration by parts:∫u dv = uv −

∫v, du

{u = f(x) v = g(x)du = f ′(x) dx dv = g′(x) dx

b.) Evaluate by using a substitution u = g(x):∫f ′(g(x))g′(x) dx =

∫f(u) du

Math 3 Calculus - Winter 2010 - Professor Trout 1

Volume of a Right Prism

Right Prism with base (cross-sectional) area A and height h: V = Ah


Cavalieri’s Principle

Bonaventura Cavalieri (1598 - 1647) stated a principle for volumes ofsolids that anticipated the integral calculus:

Suppose two solids in three dimensions are included between twoparallel planes. If every plane parallel to these two planes intersectsboth solids in cross-sections of equal area, then the two solids havethe same volumes.

NOTE: Cavalieri’s principle also holds for regions in 2D which havethe same cross-sectional lengths and, thus, have the same areas.


http://www.jimloy.com/cindy/cavalier.htm

Volume by Slicing (Loafbread)

A loaf of bread is sliced into n thin slices (of equal width ∆x) which we approx as prisms:

Volume V =n∑

i=1

V (xi) ≈n∑

i=1

A(xi)∆x


Volume by Slicing (Loafbread)

The actual volume should be the limit as n→∞ (∆x→ 0):

Volume V = limn→∞

n∑i=1

A(xi)∆x =∫ b

a

A(x) dx


Volume by Slicing (General Solid)

Suppose that a three-dimensional solid lies along the x-axis coveringthe inteval [a, b] and the cross-sectional area at x is a continuousfunction, call it it A(x). How do we define/compute it’s volume V ?

Volume V =∫ b

a

A(x) dx

NB: This works in full generality for ANY Solid Object in 3D!


http://archives.math.utk.edu/visual.calculus/5/volumes.5/index.html

Example 2

Compute the volume of a sphere of radius r = 2 at the origin by theVolume by Slicing method.

V =∫ 2

−2

A(x) dx =∫ 2

−2

π(4− x2) =323π =

43π23


Proof of Cavalieri’s Principle

Suppose two solids S1 and S2 have the same height and cross-sectional areas:

Volumes by slicing⇒ V1 =∫ b

a

A(x) dx = V2√©


Example 3 A solid object has as its base the circular region definedby the unit circle. Every cross section of the object perpendicular tothe x-axis is a triangle whose base vertices are on the circle and whoseheight equals the length of the base. Find the volume of this object.

V =∫ 1

−1

A(x) dx =∫ 1

−1

2(1− x2) dx =83


Solids of Revolution

Solids of Revolution are commonly used in engineering and manufac-turing, such as axles, funnels, pills, bottles, and pistons.



A Solid of Revolution is generated by taking a region in the plane,say the area under the graph of a function y = f(x) over [a, b],and rotating it about an axis (e.g., x-axis or another line) in threedimensions.



Every perpendicular cross-section at x is a circle of radius r = f(x),so the area function A(x) is given by:

A(x) = πr2 = π[f(x)]2.

Thus, from Volumes by Slicing, the volume V is the definite integral:

V =∫ b

a

A(x) dx =∫ b

a

π[f(x)]2 dx


Example 4: Find the volume of the solid of revolution generated byrevolving the region bounded by the x-axis, the curve y = x3−x+1and the vertical lines x = −1 and x = 1 around the x-axis.

V =∫ 1

−1

π(x3 − x+ 1)2 dx =226π105


Example 5: Find the volume of the solid of revolution generated byrevolving the region bounded by f(x) = 2−x2 and g(x) = 1 aboutthe line y = 1.

V =∫ 1

−1

A(x) dx =∫ 1

−1

π((2−x2)−1)2 dx =∫ 1

−1

π(1−x2)2 dx =16π5


Example 6: Find the volume of the solid of revolution generated byrevolving the region between the y-axis and the curve xy = 2, 1 ≤y ≤ 4, around the y-axis.

V =∫ 4

1

A(y) dy =∫ 4

1

π(r(y))2 dy = π

∫ 4

1

4y2dy = 3π


4.8: volumes of solids of revolution - dartmouth college of a right prism right prism with base...

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