# Volume: Shell Method Worksheet

Volume: Shell Method Worksheet
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1.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($x$) = $x$2 and the $x$-axis (0 ≤ $x$ ≤ 2) about $y$-axis. [Use shell method.]
 a. 4$\pi$ cubic units b. 32$\pi$ cubic units c. 8$\pi$ cubic units d. 12$\pi$ cubic units e. 16$\pi$ cubic units

#### Solution:

Equation of the curve is f(x) = x2, (0 ≤ x ≤ 2)

In the shell method, the volume of solid of revolution formed by revolving the region bounded by y = f(x) and the x - axis (axb), about y - axis is given by, V = 2π ab x · f(x) dx

Volume of solid of revolution, V = 2π 02 x (x2) dx

= 2π 02 x3 dx

= 2π [x44]02

= 2π (16 / 4) = 8π

2.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($y$) = $y$ - $y$2 and the $y$-axis (0 ≤ $y$ ≤ 1) about $x$-axis. [Use shell method.]
 a. $\frac{\pi }{6}$ cubic units b. $\frac{7\pi }{12}$ cubic units c. $\frac{\pi }{12}$ cubic units d. $\frac{\pi }{3}$ cubic units e. $\frac{14\pi }{12}$ cubic units

#### Solution:

Equation of the curve is f(y) = y - y2, (0 ≤ y ≤ 1)

In the shell method, the volume of solid of revolution formed by revolving the region bounded by x = f(y) and the y - axis (ayb), about x - axis is given by, V = 2π ab y f(y) dy

Volume of solid of revolution, V = 2π 01 y (y - y2) dy

= 2π 01 (y2 - y3) dy

= 2π [y33-y44]01

= 2π [1 / 3 - 1 / 4]

= 2π · 1 / 12 = π6 cubic units

3.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($x$) = $e$$x$, $x$-axis, $x$ = 0 and $x$ = 2 about $y$-axis. [Use shell method.]
 a. 2$\pi$ (1 - $e$2) cubic units b. $\frac{\pi }{2}$ (1 - $e$2) cubic units c. $\pi$ (1 - $e$2) cubic units d. $\pi$ (1 + $e$2) cubic units e. 2$\pi$ (1 + $e$2) cubic units

#### Solution:

Equation of the curve is f(x) = ex, (0 ≤ x ≤ 2)

In shell method, the volume of solid of revolution formed by revolving the region bounded by y = f(x) and the x - axis (axb), about y - axis is given by, V = 2π ab x f(x) dx

Volume of the solid of revolution = 2π 02 x · ex dx

= 2π [xex - ex]02

= 2π [(2e2 - e2) - (0 - 1)]

= 2π [(e2) + 1]

= 2π (1 + e2) cubic units

4.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($x$) = sin $x$, $x$-axis, $x$ = 0 and $x$ = $\frac{\pi }{2}$ about $y$-axis. [Use shell method.]
 a. 2$\pi$ cubic units b. 4$\pi$ cubic units c. $\frac{\pi }{2}$ cubic units d. $\frac{\pi }{4}$ cubic units e. $\pi$ cubic units

#### Solution:

Equation of the curve is f(x) = sin x, (0 ≤ xπ2)

In the shell method, the volume of solid of revolution is formed by revolving the region bounded by y = f(x) and the x - axis (axb), about y - axis is given by, V = 2π ab x · f(x) dx

Volume of solid of revolution, V = 2π 0π/2 x sin(x) dx

= 2π [sin x - x cos x]0π/2

= 2π(1) = 2π

5.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($y$) = $\sqrt{y}$ and the $y$-axis (0 ≤ $y$ ≤ 2) about $x$-axis. [Use shell method.]
 a. $\frac{16\pi }{5}$ cubic units b. $\frac{16\sqrt{2}\pi }{5}$ cubic units c. $\frac{8\sqrt{2}\pi }{5}$ cubic units d. $\frac{32\pi }{5}$ cubic units e. $\frac{8\pi }{5}$ cubic units

#### Solution:

Equation of the curve is f(y) = y, (0 ≤ y ≤ 2)

In the shell method, the volume of solid of revolution formed by revolving the region bounded by x = f(y) and y - axis (ayb), about x - axis is given by, V = 2π ab y f(y) dy

Volume of solid of revolution, V = 2π 02 yy dy

= 2π 02 y32 dy

= 2π [y5252]02

= 4π5(42)

= 162π5

6.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($x$) = 1 + $\frac{4}{x}$ and the $x$-axis (2 ≤ $x$ ≤ 4) about $y$-axis. [Use shell method.]
 a. 28$\pi$ b. 48$\pi$ c. 56$\pi$ d. 68$\pi$ e. 14$\pi$

#### Solution:

Equation of the curve is f(x) = 1 + 4x, (2 ≤ x ≤ 4)

In the shell method, the volume of solid of revolution formed by revolving the region bounded by y = f(x) and the x - axis (axb), about y - axis is given by, V = 2π ab x f(x) dx

Volume of solid of revolution, V = 2π 24 x (1 + 4x) dx

= 2π 24 (x + 4) dx

= 2π [x22 + 4x]24

= 2π (14)

= 28π

7.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($x$) = ln $x$, $x$-axis, $x$ = 1 and $x$ = $e$ about $y$-axis. [Use shell method.]
 a. 2$\pi$ ($e$2 - 1) cubic units b. $\pi$ ($e$2 - 1) cubic units c. 2$\pi$ ($e$2 + 1) cubic units d. $\frac{\pi }{2}\left({e}^{2}+1\right)$ cubic units e. $\frac{\pi }{2}\left({e}^{2}-1\right)$ cubic units

#### Solution:

Equation of the curve is f(x) = ln x, (1 ≤ xe)

In the shell method, the volume of solid of revolution formed by revolving to region bounded by y = f(x) and the x - axis (axb), about y - axis is given by, V = 2π ab x f(x) dx

Volume of solid of revolution, V = 2π 1e x ln x dx

= 2π [x22 (ln x - 1 / 2)]1e

= 2π [e22(1 - 1 / 2) - 1 / 2 (- 1 / 2)]

= 2π (e24 + 1 / 4)

= π2(e2+1) cubic units

8.
Find the volume of the solid of revolution formed by revolving the region bounded by $f$($y$) = $y$2 + $y$ + 1, $y$-axis, $y$ = 1 and $y$ = 2 about $x$-axis. [Use shell method.]
 a. $\frac{91\pi }{6}$ cubic units b. $\frac{41\pi }{3}$ cubic units c. $\frac{23\pi }{3}$ cubic units d. $\frac{35\pi }{3}$ cubic units e. $\frac{33\pi }{3}$ cubic units

#### Solution:

Equation of the curve is f(y) = 1 + y + y2, (1 ≤ y ≤ 2)

In the shell method, the volume of solid of revolution formed by revolving the region bounded by x = f(y) and the y - axis (axb), about x - axis is given by, V = 2π ab y f(y) dy

Volume of solid of revolution, V = 2π 12 y(1 + y + y2) dy

= 2π [y22+y33 + y44]12

= 2π (2 + 8 / 3 + 4 - 1 / 2 - 1 / 3 - 1 / 4)

= 2π (6 + 7 / 3- 3 / 4) = 91π6 cubic units.

9.
Find the volume of solid formed by revolving the region bounded by $y$ = 2$\sqrt{2x}$ and $y$ = $x$2 about $x$-axis. [Use Washer method.]
 a. $\frac{24\pi }{5}$ cubic units b. $\frac{56\pi }{5}$ cubic units c. $\frac{112\pi }{5}$ cubic units d. $\frac{48\pi }{5}$ cubic units e. $\frac{32\pi }{5}$ cubic units

#### Solution:

Equations of curves are y = 22x and y = x2

These two curves intersect at (0, 0) and (2, 4). So, x varies from 0 to 2 in the region bounded by y = 22x and y = x2.

In Washer method, the volume of solid formed by revolving the region bounded by y = f(x) and y = g(x) is given by, V = πab [(f(x))2 - (g(x))2] dx, where f(x) is the top curve and g(x) is the bottom curve.

Volume of solid of revolution, V = π0 2 [(22x)2 - (x2)2] dx

= π02 (8x - x4) dx

= π [4x2 - x55]02 = π (16 - 32 / 5) = 48π5 cubic units.

10.
Find the volume of solid formed by revolving the region bounded by $y$ = 0 and $y$ = $x$2 - 4 about $x$-axis. [Use Washer method.]
 a. $\frac{512\pi }{15}$ cubic units b. 47$\pi$ cubic units c. 51$\pi$ cubic units d. 17$\pi$ cubic units e. 85$\pi$ cubic units

#### Solution:

Equations of curves are y = 0 and y = x2 - 4

These two curves intersect at (- 2, 0) and (2, 0). So, x varies from - 2 to 2 in the region bounded by y = 0 and y = x2 - 4.

In Washer method, the volume of solid formed by revolving the region bounded by y = f(x) and y = g(x) is given by, V = πab [(f(x))2 - (g(x))2)] dx, where f(x) is the top curve and g(x) is bottom curve.

Volume of solid of revolution, V = π-22 [(x2 - 4)2 - (0)2] dx

= 2π 02 [16 + x4 - 8x2] dx

= 2π [16x + x55-8x33]02

= 2π (32 + 32 / 5 - 64 / 3) = 512π15 cubic units.