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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, $\frac{\pi }{2}$]?
 a. $\frac{{\pi }^{2}}{6}$ cubic units b. $\frac{{\pi }^{2}}{2}$ cubic units c. $\frac{{\pi }^{2}}{4}$ cubic units d. $\frac{{\pi }^{2}}{8}$ 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 $y$ = $e$$x$, $y$ = 0, $x$ = 0, and $x$ = 1.
 a. ($\frac{\pi }{2}$)$e$2 cubic units b. ($\frac{\pi }{2}$)($e$2 + 1) cubic units c. ($\frac{\pi }{2}$)($e$2 - 1) cubic units d. ($\frac{\pi }{2}$)($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 - $x$2 and $y$ = 0.
 a. 33$\pi$ cubic units b. 30$\pi$ cubic units c. 34$\pi$ cubic units d. 32$\pi$ 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 $x$ = $e$y, $x$ = 0, $y$ = 1, and $y$ = 2?
 a. ($\frac{\pi }{2}$) $e$2($e$2 + 1) b. ($\frac{\pi }{2}$)$e$4 c. ($\frac{\pi }{2}$) $e$2 ($e$2 - 1) d. ($\frac{\pi }{2}$) $e$2

#### 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 $x$ = $e$$y$, $x$ = 0, $y$ = 0, and $y$ = 1.
 a. 3$\pi$ cubic untis b. 4$\pi$ cubic untis c. 2$\pi$ cubic untis d. 5$\pi$ 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. 2$\pi$ln2 b. 2$\pi$(ln 2 - $\frac{3}{4}$) c. 2$\pi$ln 4 d. 2$\pi$(ln 4 - $\frac{3}{4}$)

#### 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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