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Volumes by Disks and Washers Volume of a cylinder A cylinder is a solid where all cross sections are the same. The volume of a cylinder is A · h where A is the area of a cross section and h is the height of the cylinder.
For a solid S for which the cross sections vary, we can approximate the volume using a Riemann sum.
The areas of the cross sections (taken perpendicular to the x-axis) of the solid shown on the left above vary as x varies. The areas of these cross sections are thus a function of x, A(x), defined on the interval [a, b]. The volume of a slice of the solid above shown in the middle picture, is approximately the volume of a cylinder with height ∆x and cross sectional area A(x∗ i ). In the picture on the right, we use 7 such slices to approximate the volume of the solid. The resulting Riemann sum is
V ≈
∑^7
i=
A(x∗ i )∆x.
The volume is the limit of such Riemann sums:
V = lim n→∞
∑^ n
i=
A(x∗ i )∆x =
∫ (^) b
a
A(x)dx.
Thus if we have values for the cross sectional area at discrete points x 0 , x 1 ,... , xn, we can estimate the volume from the data using a Riemann sum. On the other hand if we have a formula for the function A(x) for a ≤ x ≤ b, we can find the volume using the Fundamental theorem of calculus, or in the event that we cannot find an antiderivative for A(x), we can estimate the volume using a Riemann sum.
V =
∫ (^) b
a
A(x)dx.
Example The base of a solid is the region enclosed by the curve y = (^1) x and the lines y = 0, x = 1 and x = 3. Each cross section perpendicular to the x-axis is an isosceles right angled triangle with the hypotenuse across the base. Find the volume of the solid.
Example Find the volume of the solid obtained from revolving the region bounded by the curve y =
x + 1, x = 0, x = 3 and y = 0 (the x axis) about the x axis.
Method of Washers
Let f (x) and g(x) be continuous functions on the interval [a, b] with f (x) ≥ g(x) ≥ 0. Let R denote the region bounded above by y = f (x), below by y = g(x) and the lines x = a and x = b. Let S be the solid obtained by revolving the region R around the x axis. The cross sections of S are washers with area is given by
A(x) = π(outer radius)^2 − π(inner radius)^2 = π[f (x)^2 ] − π[g(x)]^2.
The volume of S is given by
V =
∫ (^) b
a
π[f (x)^2 ] − π[g(x)]^2 dx =
∫ (^) b
a
π[f (x)^2 − g(x)^2 ]dx
Example Find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y =
x and the lines x = 0 and x = 1 about the x axis. We see from the pictures below how the formula is derived:
Working with respect to the y axis
Example Let S be a solid bounded by the parallel planes perpendicular to the y axis, y = c and y = d. If for each y in the interval [c, d] the cross sectional area of S perpendicular to the y axis is A(y), the volume of the solid S is
V =
∫ (^) d
c
A(y)dy
(Provided that A(y) is an integrable function of y)
Example Find the volume of a pyramid with height 10 in. and square base whose sides have length 4 in.
Each cross section of the pyramid perpendicular to the y axis is a square. To determine the length of the side of the square at y, we consider the triangle below, bounded by the y axis, the x axis and the line along the side of the pyramid directly above the x axis. The length of the side of the cross sectional square at y is 2L and the cross sectional area at y is A(y) = 4L^2. We would like to express this in terms of y.
0 2
L
x-axis
y-axis
y
10
By simiar triangles we have (^10) L− y = 102. This gives 2(10 − y) = 10L and L = 105 − y. Therefore the cross sectional area at y is given by A(y) = 4L^2 = 254 (10 − y)^2 = 254 (100 − 20 y + y^2 ). By the formula, the volume of the pyramid is
∫ (^10)
0
(100 − 20 y + y^2 )dy =
0
(100 − 20 y + y^2 )dy =
[
100 y − 10 y^2 + y^3 / 3
] 10
0
= 160/ 3
Solids of Revolution; Revolving around the y axis
Let f (y) be a continuous function on [c, d] with f (y) ≥ 0 for all y ∈ [c, d]. Let R denote the region between the curve x = f (y) and the y-axis and the lines y = c and y = d. When the region R is revolved around the y-axis, it generates a solid with circular cross sections of radius f (y). The area of the cross section at y is the area of such a circle;
A(y) = π[f (y)]^2
and the volume of the solid (of revolution) generated by R is
V =
∫ (^) d
c
π[f (y)]^2 dy.
Example Find the volume of the solid generated by revolving the region bounded by the curve x = y^2 and the lines y = 0, y = 2 and x = 0(the y axis) about the y axis.
V =
0
πy^4 dy = π
y^5 5
2
0
= π
