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Volumes of Revolution ('discs' and 'shells') Worksheet

Volumes of Revolution ('discs' and 'shells') Worksheet
  • Page 1
 1.  
What is the volume of the solid generated by rotating about the x - axis the plane region bounded by y = cos x and y = 0 over [0, π2]?
a.
π26 cubic units
b.
π22 cubic units
c.
π24 cubic units
d.
π28 cubic units


Solution:


The plane region R bounded by the curves y = f(x) = cosx, y = 0 over [0, π2] is the shaded region shown.

The volume of the solid generated by rotating the plane region between x = 0, x = π2 around x - axis.

= V = 0π/2πy2 dx
[Disc method.]

= π0π/2cos2 x dx
[Substitute y = cos x.]

= π0π/2(1+ cos 2x2) dx
[Use cos2 x = 1+ cos 2x2.]

= π2[0 π/2 dx + 0 π/2cos 2x dx]

= π2[x |0π/2 + 1 / 2(sin 2x) |0π/2]

= π2[(π2 - 0) + 1 / 2(sin 2(π2) - sin 2(0))]

= (π2)[π2+ 1 / 2(0 - 0)]

= π24 cubic units.


Correct answer : (3)
 2.  
Find the volume of the solid formed by rotating about the x - axis the plane region bounded by y = ex, y = 0, x = 0, and x = 1.
a.
(π2)e2 cubic units
b.
(π2)(e2 + 1) cubic units
c.
(π2)(e2 - 1) cubic units
d.
(π2)(e - 1) cubic units


Solution:


The plane region bounded by the curves y = ex, y = 0, x = 0, and x = 1 is the shaded region shown.

The volume of the solid generated by rotating the shaded plane region which is in between x = 0, x = 1 around the x - axis

= V = 01π(y2) dx
[Disc method.]

=01π(ex)2 dx
[Substitute y = ex.]

= 01πe2xdx

= π01e2x dx

= π e2x2 |01

= (π2)(e2 - e0)

= (π2)(e2 - 1) cubic units.


Correct answer : (3)
 3.  
Find the approximate volume generated by rotating about the x - axis the plane region bounded by y = 4 - x2 and y = 0.
a.
33π cubic units
b.
30π cubic units
c.
34π cubic units
d.
32π cubic units


Solution:


The curve y = f(x) = 4 - x2 cuts the x - axis in two points A(- 2, 0), B(2, 0)
[Solve 4 - x2 = 0 for x-intercept.]

The curve y = f(x) = 4 - x2 cuts y - axis in C(0, 4)
[Put x = 0 in y = 4 - x2 which gives y = 4.]

The plane region bounded by y = f(x) = 4 - x2, y = 0 is the shaded region between x = - 2, x = 2 shown

The volume of the solid generated by rotating the shaded plane region which is in between x = - 2, x = 2 around the x - axis.

= V = - 22πy2 dx
[Disc method.]

= - 22π(4 - x2)2 dx
[Substitute y = 4 - x2 .]

= - 22π(16 + x4 - 8x2) dx

= π- 22(16 + x4 - 8x2) dx

= π[(16x)|- 22 + ((1 / 5)x5) |- 22 - ((8 / 3)x3) |- 22]

= π[16(2 + 2) + (1 / 5)(32 + 32) - (8 / 3)(8 + 8)]

= π[34.13]

= 34π cubic units approximately


Correct answer : (3)
 4.  
What is the volume of the solid generated by rotating about the y - axis the plane region bounded by x = ey, x = 0, y = 1, and y = 2?
a.
(π2) e2(e2 + 1)
b.
(π2)e4
c.
(π2) e2 (e2 - 1)
d.
(π2) e2


Solution:


The plane region bounded by x = ey, x = 0, y = 1, y = 2 is the shaded region shown.

The volume of the solid generated by rotating the shaded plane region between y = 1, y = 2 around the y - axis

= V = 12 π(x2) dy
[Disc Method.]

= 12π(ey)2 dy
[Substitute x = ey.]

= π12e2 y dy

= π(e2y2) |12

= (π2)[e4 - e2]

= π2e2(e2 - 1)


Correct answer : (3)
 5.  
Find the volume of the solid generated by rotating about the x - axis the region bounded by x = ey, x = 0, y = 0, and y = 1.
a.
3π cubic untis
b.
4π cubic untis
c.
2π cubic untis
d.
5π cubic untis


Solution:


The plane region bounded by x = ey, x = 0, y = 0, y = 1 is the shaded region shown.

The volume of the solid generated by rotating the shaded region between y = 0, y = 1 around x - axis

= V = 012πyx dy
[Shell method.]

= 2π01yey dy
[Substitute x = ey.]

= 2π[(yey) |01 - (ey)) |01]
[Use integration by parts.]

= 2π[(1(e) - 0) - (e - 1)]

= 2π cubic untis


Correct answer : (3)
 6.  
What is the volume of the solid generated by rotating about the x - axis the region bounded by x = ln y, x = 0, y = 1, and y = 2?
a.
2πln2
b.
2π(ln 2 - 3 4)
c.
2πln 4
d.
2π(ln 4 - 3 4)


Solution:


The plane region bounded by x = lny, x = 0, y = 1, y = 2 is the shaded region shown.

The volume of the solid generated by rotating the shaded region between y = 1, y = 2 around x - axis.

V = 122πyx dy
[Shell method.]

= 2π12y ln y dy
[Substitute x = lny.]

= 2π{[ln y(y22)] |12 - 121y(y22)]}dy
[Use integration by parts.]

= 2π [2ln 2 - 1 / 4(y2) |12]

= 2π [2ln 2 - 3 / 4]

= 2π[ln 4 - 3 / 4]


Correct answer : (4)
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