2.2 Solids of Revolution

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We use disks, washers and shells to find the volume of a solid of revolution.

A special type of solid whose cross-sections are familiar geometric shapes is the solid of revolution. We will use the idea of the last section, namely that volume can be found by integrating cross-sectional area, to find the volume of such a solid.

Solid of revolution A solid of revolution is a solid obtained by revolving a region in the $xy$ -plane about a line called the axis of revolution.

Cylinders, cones and spheres are examples of well known solids which are solids of revolution. For simplicity, we will only consider horizontal and vertical lines for our axis of revolution. Furthermore, we will assume that our region in the plane is bounded by functions of $x$ . In the event that we have a region bounded by functions of $y$ , we can interchange the variables $x$ and $y$ throughout the problem to convert the problem to the setting of functions of $x$ .

The main formulas we will need from geometry are the area of a disk, area of a washer and the lateral surface area of a cylinder:

To see how these particular shapes come into play, imagine revolving a vertical segment about the axis of revolution. If the axis of revolution is horizontal and the vertical segment touches the axis at one endpoint, this will produce a disk (see the figure below).

If the vertical segment does not touch the horizontal axis of revolution, then revolving it will create a washer.

Finally, if the axis of revolution is vertical, then revolving the vertical segment about the axis of revolution will produce the lateral surface of a cylinder (see the figure below).

1 Disks

example 1 Find the volume of a right, circular cone with height 4, and base radius 2.
This cone can be obtained by revolving the the region between the line $y=x/2$ and the $x$ -axis from $x = 0$ to $x = 4$ about the $x$ -axis.
To begin, we sketch the region (see below) and we draw a vertical line segment in the region at $x$ . Revolving this segment about the axis of revolution (the $x$ -axis) creates a disk. The area of this disk is $A = \pi r^2,$ where the radius, $r$ , is given by $r = \frac {x}{2} - 0 = \frac {x}{2}.$ We find the volume by integrating area: $V = \int _a^b \text {(area)} \; dx = \int _0^4 \pi r^2 \, dx = \int _0^4 \pi \left (\frac {x}{2}\right )^2 \, dx.$ Computing this integral gives $V = \frac {\pi }{4} \int _0^4 x^2 \, dx = \frac {\pi }{4} \cdot \frac {x^3}{3} \bigg |_0^4 = \frac {\pi }{4}\cdot \frac {64}{3} = \frac {16\pi }{3}.$

(problem 1a) Find the formula for the volume of a right, circular cone with height $5$ , and base radius $3$ . Include a rough sketch of the region being revolved.

To create the cone, consider revolving a line through the origin about the $x$ -axis.
The equation of this line is $y = \answer {\frac {3}{5} x}$ .
We will integrate over the interval $\left (\answer {0},\answer {5}\right )$

The definite integral that gives the volume of the cone is:

$\displaystyle {V = \int _0^5 \pi \cdot \frac 35 x^2\; dx}$ $\displaystyle {V = \int _0^5 2\pi \left (\frac 35 x\right )^2\; dx}$ $\displaystyle {V = \int _0^5 \pi \left (\frac 35 x\right )^2\; dx}$

The volume of the cone is: $V = \answer {15 \pi }$ .

(problem 1b) Find the formula for the volume of a right, circular cone with height $H$ , and base radius $R$ . Include a rough sketch of the region being revolved.

To create the cone, consider revolving a line through the origin about the $x$ -axis.
The equation of this line is $y = \answer {\frac {R}{H} x}$ .
We will integrate over the interval $\left (\answer {0},\answer {H}\right )$

The definite integral that gives the volume of the cone is:

$\displaystyle {V = \int _0^H \pi \left (\frac {R}{H}x\right )^2\; dx}$ $\displaystyle {V = \int _0^H 2\pi \left (\frac {R}{H}x\right )^2\; dx}$ $\displaystyle {V = \int _0^H \pi \left (\frac {R}{H}\right )x^2\; dx}$

The formula for the volume of a cone is: $V = \answer {\pi R^2 H/3}$ .

(problem 1c) Find the formula for the volume of the frustum of the right, circular cone obtained by revolving the region bounded by the lines $y = 2x+1, y=0, x = 0,$ and $x = 2$ about the $x$ -axis. Include a rough sketch of the region being revolved.

The radius of the disk at $x$ is $r = \answer {2x+1}$
We will integrate over the interval $\left (\answer {0},\answer {2}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = \pi \int _{0}^2 \left (4x^2 +1\right ) \; dx}$ $\displaystyle {V = \pi \int _{0}^2 \left (4x^2 + 4x +1\right ) \; dx}$ $\displaystyle {V = \pi \int _{0}^2 \left (2x^2 + 4x +1\right ) \; dx}$

The volume of the solid is $V = \answer {\frac {62}{3} \pi }$ .

example 2 Find the volume of a sphere of radius 2.
We can think of this sphere as the solid of revolution obtained by revolving the region between the semi-circle $y = \sqrt {4-x^2}$ and the $x$ -axis about the $x$ -axis. To begin, we sketch the region including a vertical line segment in the region at $x$ (see below). Revolving this segment about the axis of revolution (the $x$ -axis) creates a disk. The area of this disk is $A = \pi r^2,$ where the radius, $r$ , is given by $r = \sqrt {4 - x^2} - 0 = \sqrt {4-x^2}.$ We find the volume by integrating area: $V = \int _a^b \text {(area)} \; dx = \int _{-2}^2 \pi r^2 \, dx = \int _{-2}^2 \pi \left (\sqrt {4-x^2}\right )^2 \, dx.$ Computing this integral gives $V = \pi \int _{-2}^2 (4 - x^2) \, dx = \pi \left (4x - \frac {x^3}{3}\right ) \bigg |_{-2}^2$ $= \pi \left [\left (8 - \frac 83 \right ) - \left (-8 + \frac 83\right )\right ] = \pi \left (16 - \frac {16}{3}\right ) = \frac {32 \pi }{3}$

(problem 2a) Find the formula for the volume of a sphere of radius $3$ . Include a rough sketch of the region being revolved.

To create the sphere, consider revolving a semi-circular region centered at the origin about the $x$ -axis.
The equation of this semi-circle is $y = \answer {\sqrt {9-x^2}}$ .
We will integrate over the interval $\left (\answer {-3},\answer {3}\right )$

The definite integral that gives the volume of the sphere is:

$\displaystyle {V = \int _{-3}^3 \pi \sqrt {9 - x^2} \; dx}$ $\displaystyle {V = \int _{-3}^3 \pi \left (\sqrt {9-x^2}\right )^2\; dx}$ $\displaystyle {V = \int _{-3}^3 2\pi \left (\sqrt {9-x^2}\right )^2\; dx}$

The volume of the sphere is: $V = \answer {36 \pi }$ .

(problem 2b) Find the formula for the volume of a sphere of radius $R$ . Include a rough sketch of the region being revolved.

To create the sphere, consider revolving a semi-circular region centered at the origin about the $x$ -axis.
The equation of this semi-circle is $y = \answer {\sqrt {R^2-x^2}}$ .
We will integrate over the interval $\left (\answer {-R},\answer {R}\right )$

The definite integral that gives the volume of the sphere is:

$\displaystyle {V = \int _{-R}^R\pi \left (\sqrt {R^2-x^2}\right )^2\; dx}$ $\displaystyle {V = \int _{-R}^R \pi \sqrt {R^2 - x^2} \; dx}$ $\displaystyle {V = \int _{-R}^R 2\pi \left (\sqrt {R^2-x^2}\right )^2\; dx}$

The formula for the volume of a sphere with radius $R$ is: $V = \answer {4/3 \pi R^3 }$ .

example 3 Find the volume of the solid obtained by revolving the region bounded by the graphs of $y=x^2$ and $y = 4$ about the line $y = 4$ .
We first find the points of intersection of the parabola $y=x^2$ and the horizontal line $y = 4$ : $x^2 = 4 \implies x = \pm 2$ Next, we sketch the region including a vertical line segment in the region at $x$ (see below).

Revolving this segment about the line $y = 4$ creates a disk (which is a cross-section of the solid). The area of this disk is $A = \pi r^2,$ where the radius is given by $r = 4-x^2.$

We find the volume by integrating the area of a cross-section: $V = \int _a^b \text {(area)} \; dx = \int _{-2}^2 \pi r^2 \, dx = \int _{-2}^2 \pi \left (4 - x^2\right )^2 \, dx.$ Computing this integral gives $V = \pi \int _{-2}^2 (16 - 8x^2 + x^4) \, dx = \pi \left (16x - \frac {8x^3}{3} + \frac {x^5}{5}\right ) \bigg |_{-2}^2$ $= \pi \left [\left (32 - \frac {64}{3} + \frac {32}{5}\right ) - \left (-32 + \frac {64}{3} - \frac {32}{5}\right )\right ] = \pi \left (64 - \frac {128}{3} + \frac {64}{5}\right ) = \frac {512 \pi }{15}$

(problem 3a) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y=x^2$ and $y=1$ about the line $y = 1$ . Include a rough sketch of the region being revolved.

The radius of the disk at $x$ is $r = \answer {1-x^2}$
We will integrate over the interval $\left (\answer {-1},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = \int _{-1}^1 \pi \left (1-x^2\right )^2\; dx}$ $\displaystyle {V = \int _{-1}^1 \pi (1-x^2)\; dx}$ $\displaystyle {V = \int _{-1}^1 2\pi \left (1-x^2\right )^2\; dx}$

The volume of the solid is $V = \answer {\frac {16}{15} \pi }$ .

(problem 3b) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y=\sqrt x, y=1$ and $x = 0$ about the line $y=1$ . Include a rough sketch of the region being revolved.

The radius of the disk at $x$ is $r = \answer {1-\sqrt x}$
We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = \int _0^1 \pi (1-\sqrt x) \; dx}$ $\displaystyle {V = \int _0^1 2\pi \left (1-\sqrt x\right )^2\; dx}$ $\displaystyle {V = \int _0^1 \pi \left (1-\sqrt x\right )^2\; dx}$

The volume of the solid is $V = \answer {\frac 16 \pi }$ .

2 Washers

In the next series of examples, the region will not border the axis of revolution. In this situation, the resulting cross-sections are washers rather than disks. The area of a washer with inner radius, $r$ , and outer radius, $R$ , is given by $A = \pi (R^2 - r^2).$

We now use this to calculate volumes of solids of revolutions about the $x$ -axis in which the region does not border the axis.

example 4 Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = e^x, y = 3$ , and $x = 0$ about the $x$ -axis.
First, note that the curves $y = e^x$ and $y = 3$ intersect at $x = \ln (3)$ .

Thus, to compute the volume of the resulting solid, fix a value of $x$ between $0$ and $\ln (3)$ and consider a vertical segment in the region at this $x$ -value.

Revolving this individual segment about the $x$ -axis yields a washer with inner radius, $r = e^x$ , outer radius, $R = 3$ .

The area of this washer is then $A = \pi (R^2 - r^2) = \pi \left [3^2 - \left (e^x\right )^2\right ] = \pi \left (9 - e^{2x}\right ) .$ The volume of the solid can be obtained by integrating the area of a cross-section as $x$ ranges from $0$ to $\ln 3$ : $V = \int _a^b (area) \; dx = \int _0^{\ln 3} \pi \left (9 - e^{2x}\right ) \; dx = \pi \left (9x - \frac {e^{2x}}{2}\right )\bigg |_0^{\ln 3} = \left (9\ln (3)-4\right )\pi .$

(problem 4) Find the volume of the solid obtained by revolving the region between $y= \frac {1}{x}$ and $y = 1$ from $x = 1$ to $x = 2$ about the $x$ -axis. Include a rough sketch of the region being revolved.

The outer radius of the washer at $x$ is $R = \answer {1}$
The inner radius of the washer at $x$ is $r = \answer {1/x}$

We will integrate over the interval $\left (\answer {1},\answer {2}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = \int _{1}^2 \pi \left (1-\frac {1}{x^2}\right )\; dx}$ $\displaystyle {V = \int _{1}^2 \pi \left (1-\frac {1}{x}\right )\; dx}$ $\displaystyle {V = \int _{1}^2 \pi \left (1-\frac {2}{x} +\frac {1}{x^2}\right )\; dx}$

The volume of the solid is $V = \answer {\pi /2 }$ .

example 5 Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = \sec (x), y = 2, x = 0$ , and $x = \frac {\pi }{4}$ about the $x$ -axis.
We begin with a sketch of the region.

To compute the volume of the resulting solid, fix a value of $x$ between 0 and $\frac {\pi }{4}$ and consider a vertical segment in the region at this $x$ -value.

Revolving this individual segment about the $x$ -axis yields a washer with inner radius, $r = \sec (x)$ and outer radius, $R = 2$ .

The area of this washer is then $A = \pi (R^2 - r^2) = \pi \left [2^2 - \sec ^2(x)\right ] = \pi \left [4 - \sec ^2(x)\right ] .$ The volume of the solid can be obtained by integrating the area of a cross-section as $x$ ranges from 0 to $\pi /4$ : $V = \int _a^b (area) \; dx = \int _0^{\pi /4} \pi \left [4-\sec ^2(x)\right ] \; dx = \pi \left [4x - \tan (x)\right ]\bigg |_0^{\pi /4} = \pi ^2 - \pi .$

(problem 5) Find the volume of the solid obtained by revolving the region between $y= \sqrt [3] x$ and $y = 1$ from $x = 0$ to $x = 1$ about the $x$ -axis. Include a rough sketch of the region being revolved.

The outer radius of the washer at $x$ is $R = \answer {1}$
The inner radius of the washer at $x$ is $r = \answer {x^{1/3}}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = \int _{0}^1 \pi \left (1-x^{1/6}\right )\; dx}$ $\displaystyle {V = \int _{0}^1 \pi \left (1-x^{1/3}\right )\; dx}$ $\displaystyle {V = \int _{0}^1 \pi \left (1-x^{2/3}\right )\; dx}$

The volume of the solid is $V = \answer {2\pi /5 }$ .

example 6 Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x^2$ and $y = \sqrt x$ about the line $y = 1$ .
The $x$ -coordinates of the points of intersection of the curves are: $x^2 = \sqrt x \implies x = 0 \;\; \text {and} \;\; x=1$ Next we make a sketch of the region including a vertical segment at $x$ (see below).

Revolving this vertical segment about the axis of revolution makes a washer.

The area of this cross-sectional washer is $A = \pi (R^2 -r^2),$ where $R = 1-x^2 \;\; \text {and} \;\; r = 1 - \sqrt x$ Squaring these radii gives $R^2 = (1-x^2)^2 = 1 - 2x^2 + x^4$ and $r^2 = (1-\sqrt x)^2 = 1 - 2\sqrt x + x$ The volume of the solid is found by integrating the area of this cross-sectional washer: $V = \int _a^b (\text {area}) \; dx = \int _0^1 \pi (R^2 -r^2) \; dx = \int _0^1 \pi [(1-2x^2 + x^4) - (1-2\sqrt x + x)] \; dx$

Computing this integral gives,

$V = \pi \int _0^1 (x^4 -2x^2 - x + 2\sqrt x) \; dx = \pi \left (\frac {x^5}{5} - \frac {2x^3}{3} - \frac {x^2}{2} + \frac {4x^{3/2}}{3} \right ) \bigg |_0^1$ $= \pi \left (\frac 15 -\frac 23 - \frac 12 + \frac 43 \right ) = \frac {11\pi }{30}.$

(problem 6) Find the volume of the solid obtained by revolving the region between $y= x$ and $y = x^2$ from $x = 0$ to $x = 1$ about the line $y = 1$ . Include a rough sketch of the region being revolved.

The outer radius of the washer at $x$ is $R = \answer {1-x^2}$
The inner radius of the washer at $x$ is $r = \answer {1-x}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = \int _{0}^1 \pi \left (x^4 -x^2 + 2x\right )\; dx}$ $\displaystyle {V = \int _{0}^1 \pi \left (x^4 -3x^2 + 2x\right )\; dx}$ $\displaystyle {V = \int _{0}^1 \pi \left (x^4 -3x^2 - 2x\right )\; dx}$

The volume of the solid is $V = \answer {\pi /5 }$ .

3 Cylindrical shells

We now find the volume of solids of revolution obtained by revolving a region about a vertical axis.

example 7 Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x^2, y=0$ , and $x = 1$ about the $y$ -axis.
We begin with a sketch of the region.

To compute the volume of the resulting solid, fix a value of $x$ between 0 and 1 and consider a vertical segment in the region at this $x$ -value.

Revolving this individual segment about the $y$ -axis yields a cylindrical shell with radius, $r = x,$ height, $h = x^2$ , and thickness, $dx$ .

The surface area of this shell is: $A = 2\pi rh = 2\pi x(x^2) = 2\pi x^3 .$ The volume of the solid can be obtained by integrating the area of the shells as $x$ ranges from 0 to 1: $V = \int _a^b (area) \; dx = \int _0^1 2\pi x^3 \; dx = 2\pi \frac {x^4}{4}\bigg |_0^1 = \frac {\pi }{2}.$

(problem 7) Find the volume of the solid obtained by revolving the region between $y= x+1, y = 0, x = 0$ and $x = 1$ about the $y$ -axis. Include a rough sketch of the region being revolved.

The radius of the cylindrical shell at $x$ is $r = \answer {x}$
The height of the cylindrical shell at $x$ is $h = \answer {x+1}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = 2\pi \int _{0}^1 \left (x^2 +1\right )\; dx}$ $\displaystyle {V = 2\pi \int _{0}^1 \left (x^2 +x\right )\; dx}$ $\displaystyle {V = \pi \int _{0}^1 \left (x^2 +x\right )\; dx}$

The volume of the solid is $V = \answer {5\pi /3 }$ .

example 8 Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = \sqrt x, y=2, x=0$ , and $x = 1$ about the $y$ -axis.
We begin with a sketch of the region.

To compute the volume of the resulting solid, fix a value of $x$ between 0 and 1 and consider a vertical segment in the region at this $x$ -value.

Revolving this individual segment about the $y$ -axis yields a cylindrical shell with radius, $r = x$ and height, $h = 2 - \sqrt x$ .

The surface area of the shell is: $A = 2\pi rh = 2\pi x(2 - \sqrt x) = 2\pi \left (2x - x^{3/2}\right ) .$ The volume of the solid can be obtained by integrating the area of the shells as $x$ ranges from 0 to 1: $V = \int _a^b (area) \; dx = \int _0^1 2\pi \left (2x - x^{3/2}\right ) \; dx = 2\pi \left (x^2 - \frac 25 x^{5/2}\right )\bigg |_0^1 = \frac {6\pi }{5}.$

(problem 8) Find the volume of the solid obtained by revolving the region between $y= x$ and $y=x^2$ about the $y$ -axis. Include a rough sketch of the region being revolved.

The radius of the cylindrical shell at $x$ is $r = \answer {x}$
The height of the cylindrical shell at $x$ is $h = \answer {x-x^2}$

We will integrate over the interval $\left (\answer {0},\answer {1}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = 2\pi \int _{0}^1 \left (x^2 -x^3\right )\; dx}$ $\displaystyle {V = 2\pi \int _{0}^1 \left (x^2 +x^3\right )\; dx}$ $\displaystyle {V = 2\pi \int _{0}^1 \left (x^4 -x^6\right )\; dx}$

The volume of the solid is $V = \answer {\pi /6 }$ .

example 9 Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x-x^2$ and $y=0$ about the line $x = -1$ .
We begin with a sketch of the region.

Note that the parabola $y=x-x^2$ crosses the $x$ -axis at $x = 0$ and $x = 1$ . To compute the volume of the resulting solid, fix a value of $x$ between 0 and 1 and consider a vertical segment in the region at this $x$ -value.

Revolving this individual segment about the line $x = -1$ yields a cylindrical shell with radius, $r = x+1$ and height, $h = x-x^2$ .

The surface area of the shell is: $A = 2\pi rh = 2\pi (x+1)(x-x^2) = 2\pi \left (x -x^3 \right ) .$ The volume of the solid can be obtained by integrating the area of the shells as $x$ ranges from 0 to 1: $V = \int _a^b (area) \; dx = \int _0^1 2\pi \left (x - x^3\right ) \; dx = 2\pi \left (\frac {x^2}{2} - \frac {x^4}{4}\right )\bigg |_0^1 = \frac {\pi }{2}.$

(problem 9) Find the volume of the solid obtained by revolving the region between $y= \frac {1}{x}, y = 0, x = 1$ and $x=2$ about the line $x = 1$ . Include a rough sketch of the region being revolved.

The radius of the cylindrical shell at $x$ is $r = \answer {x-1}$
The height of the cylindrical shell at $x$ is $h = \answer {1/x}$

We will integrate over the interval $\left (\answer {1},\answer {2}\right )$

The definite integral that gives the volume of the solid is:

$\displaystyle {V = 2\pi \int _{1}^2 \left (x - \frac {1}{x} \right )\; dx}$ $\displaystyle {V = 2\pi \int _{1}^2 \left (1 - \frac {x}{x} \right )\; dx}$ $\displaystyle {V = 2\pi \int _{1}^2 \left (1-\frac {1}{x}\right )\; dx}$

The volume of the solid is $V = \answer {2\pi (1-\ln (2)) }$ .

Here is a video on the method of washers and the method of cylindrical shells

4 More Problems

(problem 10) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x$ and $y = \sqrt x$ about the line $x = -1$ . Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$ .
The radius and height as functions of $x$ are:
$r = \answer {x+1} \;\;\;$ and $\; h = \answer {\sqrt {x} - x}$ .

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _0^1 2\pi \left (-x^2 + x^{3/2} - x + x^{1/2}\right )\; dx}$ $\displaystyle {V = \int _0^1 2\pi \left (x^2 + x^{3/2} - x - x^{1/2}\right )\; dx}$ $\displaystyle {V = \int _0^1 2\pi \left (-x^2 + x^{3/2} + x - x^{1/2}\right )\; dx}$

The volume of the solid is $V = \answer {7\pi /15}$ .

(problem 11) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x$ and $y = \sqrt x$ about the line $y = 1$ . Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$ .
The outer radius, $R$ , and inner radius, $r$ , as functions of $x$ are:
$R = \answer {1-x} \;\;\;$ and $\; r = \answer {1-\sqrt {x}}$ .

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _0^1 \pi \left (x^2 - x + 2\sqrt x\right )\; dx}$ $\displaystyle {V = \int _0^1 \pi \left (x^2 - 3x + 2\sqrt x \right )\; dx}$ $\displaystyle {V = \int _0^1 \pi \left (x^2 -3x - 2\sqrt x\right )\; dx}$

The volume of the solid is $V = \answer {\pi /6}$ .

(problem 12) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = \frac {2}{x}, y = 0, x = 1$ , and $x = 2$ about the $x$ -axis. Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {1}\;\;\;$ and $\;\;\; b = \answer {2}$ .
The radius, $r$ , as a function of $x$ is:
$r = \answer {2/x}.$

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _1^2 2\pi \frac {2}{x}\; dx}$ $\displaystyle {V = \int _1^2 \pi \frac {2}{x^2}\; dx}$ $\displaystyle {V = \int _1^2 \pi \frac {4}{x^2}\; dx}$

The volume of the solid is $V = \answer {2\pi }$ .

(problem 13) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x^3, y = 0$ , and $x = 1$ about the $x$ -axis. Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$ .
The radius, $r$ , as a function of $x$ is:
$r = \answer {x^3}.$

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _0^1 \pi x^3\; dx}$ $\displaystyle {V = \int _0^1 \pi x^5\; dx}$ $\displaystyle {V = \int _0^1 \pi x^6\; dx}$

The volume of the solid is $V = \answer {\pi /7}$ .

(problem 14) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x^2, y = 0, x = 0$ and $x = 2$ about the line $y = 4$ . Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {2}$ .
The outer radius, $R$ , and inner radius, $r$ , as functions of $x$ are:
$R = \answer {4} \;\;\;$ and $\; r = \answer {4-x^2}$ .

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _0^2 \pi \left (2x^2 - x^4\right )\; dx}$ $\displaystyle {V = \int _0^2 \pi \left (x^4 - 8x^2 \right )\; dx}$ $\displaystyle {V = \int _0^2 \pi \left (8x^2 - x^4\right )\; dx}$

The volume of the solid is $V = \answer {224\pi /15}$ .

(problem 15) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = 2e^x, y = 0, x = -1$ , and $x = \ln 2$ about the $x$ -axis. Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {-1}\;\;\;$ and $\;\;\; b = \answer {\ln 2}$ .
The radius, $r$ , as a function of $x$ is:
$r = \answer {2e^x}.$

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _{-1}^{\ln 2} 4\pi e^x\; dx}$ $\displaystyle {V = \int _{-1}^{\ln 2} 2\pi e^{2x}\; dx}$ $\displaystyle {V = \int _{-1}^{\ln 2} 4\pi e^{2x}\; dx}$

The volume of the solid is $V = \answer {(8-2/{e^2})\pi }$ .

(problem 16) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x^2$ and $y = x^3$ about the $y$ -axis. Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {1}$ .
The radius and height as functions of $x$ are:
$r = \answer {x} \;\;\;$ and $\; h = \answer {x^2 -x^3}$ .

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _0^1 2\pi \left (x^3 -x^4\right )\; dx}$ $\displaystyle {V = \int _0^1 2\pi \left (x^4 -x^3\right )\; dx}$ $\displaystyle {V = \int _0^1 2\pi \left (x^2 -x^3\right )\; dx}$

The volume of the solid is $V = \answer {\pi /10}$ .

(problem 17) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = x-1, y = 2- x$ and $x = 0$ about the line $y = -1$ . Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {3/2}$ .
The outer radius, $R$ , and inner radius, $r$ , as functions of $x$ are:
$R = \answer {3-x} \;\;\;$ and $\; r = \answer {x}$ .

The definite integral that represents the volume is given by:

$\displaystyle {V = \int _0^{3/2} \pi \left (6x- 9\right )\; dx}$ $\displaystyle {V = \int _0^{3/2} \pi \left (9 - 6x \right )\; dx}$ $\displaystyle {V = \int _0^{3/2} \pi \left (x^2 -6x + 9\right )\; dx}$

The volume of the solid is $V = \answer {27\pi /4}$ .

(problem 18) Find the volume of the solid obtained by revolving the region bounded by the graphs of $y = \sqrt [4] x, y = 0$ and $x = 16$ about the line $x = 16$ . Include a rough sketch of the region being revolved.
The method we will use is:

Disks Washers Shells

The formula associated with this method is:

$\displaystyle {V = \int _a^b \pi r^2 \, dx}$ $\displaystyle {V = \int _a^b \pi (R^2 - r^2) \, dx}$ $\displaystyle {V = \int _a^b 2\pi rh \, dx}$

The left and right endpoints of integration are:
$a = \answer {0}\;\;\;$ and $\;\;\; b = \answer {16}$ .
The radius and height as functions of $x$ are:
$r = \answer {16-x} \;\;\;$ and $\; h = \answer {x^{1/4} }$ .

The definite integral that represents the volume is given by:

$\displaystyle {V = 2\pi \int _0^{16} \left (16x^{1/4} - x\right )\; dx}$ $\displaystyle {V = 2\pi \int _0^{16} \left (16x^{1/4} - x^{5/4} \right )\; dx}$ $\displaystyle {V = 2\pi \int _0^{16} \left (16x^{1/4} + x - 16x^{5/4}\right )\; dx}$

The volume of the solid is $V = \answer {16384\pi /45}$ .

2024-09-27 13:59:01