Integration - Volume of Revolution

Suppose that f(x) is a function of x and the curve of the function is rotated around the x-axis through 3600. It clearly forms a solid object. The volume of the object can be calculated as follows:

Volume of revolution

Consider a small disk with thickness dx. If its volume is dv,
dv = ∏ y2 dx
dv = ∏ f(x)2 dx
In order to find the total volume, V of the object, let's integrate the above between a and b.
V = ∏ ab f(x)2 dx.

So, in order to find the volume of revolution:
1) Square the function and then integrate it
2) Then, multiply the answer by ∏

E.g.1

Show that the volume of a cylinder is, V = ∏ r2 h.

A cylinder is produced when the line y = r is rotated around x-axis.

Volume of cylinder

V = ∏ 0h f(x)2 dx
V = ∏ 0h r2 dx
V = ∏ [r2x]h0
V = ∏ r2 h.

E.g.2

Show that the volume of a cone is, V = 1/3 ∏ r2 h.

A cone is produced when a line y = mx is rotated around the x-axis.

Volume of cylinder

V = ∏ 0h f(x)2 dx
V = ∏ 0h (mx)2 dx
V = ∏ m2 0h x2 dx
V = ∏ m2 [ x3 / 3]0h
m, the gradient of the line, = r /h
V = ∏ r2 / h2 h3 / 3
V = 1/3 ∏ r2 h

E.g.3

Show that the volume of a sphere is, V = 4/3 ∏ r3.

A sphere is produced when the semi-circle, x2 + y2 = r2 is rotated around the x-axis.

Volume of cylinder

V = ∏ -rr f(x)2 dx
V = ∏ -rr [ √r2 - x2 ]2 dx
V = ∏ -rr[r2 - x2] dx
V = ∏ [ r2x - x3/3 ]r-r
V = ∏ [ r3 - r3/3] - [-r3 - (-r)3/3 ]
V = ∏ [ r3 - r3/3 + r3 - r3/3 ]
V = ∏ [ 2r3 - 2r3 / 3 ]
V = 4/3 ∏ r3

E.g.4

Find the volume of the object formed when the curve, y = x2 + 2 around the x-axis, between x = 0 and x = 3.

A 'vase' is produced when the curve, y = x2 + 2 is rotated around the x-axis.

Volume of cylinder

V = ∏ 03 f(x)2 dx
V = ∏ 03 [ x2 + 2 ]2 dx
V = ∏ 03[x4 + 4x2 + 4] dx
V = ∏ [ x5/5 + 4x3/3 + 4x ]30
V = ∏ [ 35/5 + 4(3)3/3 + 4(3)] - [05/5 + 4(0)0/3 + 4(0) ]
V = ∏ [96.6]
V = 96.6∏
V = 96.6∏



 

 

 

 

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