Introduction
In this section we will discuss about application of integral to compute the volume of solids: numerical measure of a three-dimensional solid.Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: V=lbh. The formulas for the volumes of a sphere is \(V=\frac{4}{3} \pi r^3\), a cone \(V=\frac{1}{3} \pi r^2 h\), a pyramid \(V=\frac{1}{3} A h \) have also been introduced. Although some of these formulas were derived using geometry alone, all these formulas can be obtained by using integration.
Three are four common approach of computing volume of objects using integral, these are calculating volumes of solids using methods of
- cross-section (slicing method)
- disk
- washers and
- shells
cross-section (the slicing method)
A cross-section of a solid is the region obtained by intersecting the solid with a plane. We can determine the volume of a solid by integrating a cross-section (the slicing method).
Given the cross sectional area A(x) of the solid, within the interval [[a,b], and cross sections are perpendicular to the x-axis, the volume of the solid is
\(V = \int_a^b A(x) dx\) where A(x) is the ea of cross section, dx is the thickness of the cross section.
Solved Examples
- Find the volume of solid whose base is bounded by the graph of \(f(x)=\sqrt{\sin x} \) from x=0 to x=π, and the x-axis, with perpendicular cross sections that are squares.
Solution
Volume =Area x slices
or Volume =\(\displaystyle \int_0^\pi Adx \)
or Volume =\(\displaystyle \int_0^\pi l^2 dx \)
or Volume =\(\displaystyle \int_0^\pi \sin x dx \) -
Find the volume of solid whose base is bounded by graphs of \(y=4x\) and \(y=x^2\), with perpendicular cross sections that are circles.
Drag the red point
Solution
Volume =Area x slices
or Volume =\(\displaystyle \int_0^4 Adx \)
or Volume =\(\displaystyle \int_0^4 (\pi r^2) dx \)
or Volume =\(\displaystyle \pi \int_0^4 \left (\frac{4x-x^2}{2} \right )^2 dx \)
or Volume =\(\displaystyle \frac{\pi}{4} \int_0^4 (4x-x^2)^2 dx \) -
Find the volume of solid whose base is bounded by the graph of \(x^2+y^2=9\) with perpendicular cross sections that are equilateral triangles
Drag the red point
Solution
Volume =Area x slices
or Volume =\(\displaystyle \int_{-3}^{3} \frac{\sqrt{3}a^2}{4}dx \)
or Volume =\(\displaystyle \frac{\sqrt{3}}{4} \int_{-3}^{3} a^2dx \)
or Volume =\(\displaystyle \frac{\sqrt{3}}{4} \int_{-3}^{3} \left ( 2 \sqrt{9-x^2}\right )^2 dx \)
or Volume =\(\displaystyle \sqrt{3} \int_{-3}^{3} (9-x^2) dx \) -
Find the volume of solid whose base is bounded by \(y=x^3,x=2 \), and the x-axis, and whose cross sections are perpendicular to the y-axis and are isosceles right triangles with a leg on the base of the solid.
Solution
Volume =Area x slices
or Volume =\(\displaystyle \int_0^8 \frac{1}{2} (2-\sqrt[3]{y})^2 dy \) - The base of a solid is the region between \(f(x)=x^2-1, g(x)=1-x^2\), and its cross-sections perpendicular to the x -axis are equilateral triangles, find the volume of the solid so formed.
disk method
We can get the volume of a solid by rotating the area around a line, which results in a right cylinder, or a disk.We know that, the formula for the volume of a cylinder is
Volume= πr2⋅height
in which the radius is the distance from the axis of revolution to the function, and the “height” of each disk, or slice is “dx”
This formula is applied when a bounded region is rotated about x–axis or about y–axis, and a solid is formed, the formula is
\(\displaystyle \int_a^b \pi r^2 dx\)
where r, is the radius of disk, \(A(x)=\pi r^2\) is the area of disk, dx is the thickness of the disk.
We can find volume of such solid by rotating y=f(x) about x–axis by
\(V = \int_a^b \pi y^2 dx \)
where, \(A = \pi y^2\) is cross sectional area obtained by slicing.
We can find volume of such solid by rotating y=f(x) about y–axis by
\(V = \int_a^b \pi x^2 dy \)
where, \(A = \pi x^2\) is cross sectional area obtained by slicing.
Solved Examples
- Find the volume of a pyramid that is 20 metres tall with a square base 20 metres on a side.
- Determine the volume of the solid obtained by rotating the region bounded by \(y=16-x^2,y=0\) about the x-axis
Solution
The first thing to do is get a sketch of the bounding region and the solid obtained by rotating the region about the x -axis.
Here are the graphs.The cross-sectional area is,
A(x)=π r2
or A(x)= π(16- x2)2
Next, we need to determine the limits of integration.
The limits are x=-4 and x=4.
Thus the volume of this solid is
\(V=\int_{-4}^4 A(x) dx\)
or \(V=\pi \int_{-4}^4 (16-x^2)^2 dx\) - Determine the volume of the solid obtained by rotating the region bounded by \(y=x^2-4x+5,x=1,x=4\) and the x-axis about the x-axis
Solution
The first thing to do is get a sketch of the bounding region and the solid obtained by rotating the region about the x -axis.
Here are the graphs.The cross-sectional area is,
A(x)= \( \pi ( x^2- 4x + 5)^2\)
Next, we need to determine the limits of integration.
The limits are x=1 and x=4.
Thus the volume of this solid is
\(V=\int_1^4 A(x) dx\)
or \(V= \int_1^4 \pi ( x^2- 4x + 5)^2 dx\)
or \(V=\frac{78 \pi}{5}\)
washer method
The washer method is similar to the disk method, but it covers solids of revolution with “holes” made by inner and outer functions. In this method, we need to find inner and outer radii.
Then we subtract the area of the inner solid from the area of the outer solid. Please note that, to apply this method, the inner function must be completely inside (or touching) the outer function over the integration interval.
Suppose f(x) and g(x) are non-negative and continuous on the interval [a,b] with \(f(x)\ge g(x)\). Let S be the area bounded above by f(x) and below by g(x) as well as the lines
x=a and x=b. Then the volume V formed by rotating S about the x-axis is
\(V= \pi \int_a^b [R(x)^2-r(x)^2] dx \)
where x is the location of the washer,R(x) is the radius of the outer disk, r(x) is the radius of the inner disk, and dx is the thickness of the washer.
Solved Examples
- Determine the volume of the solid obtained by rotating the portion of the region bounded by \( y=\sqrt[3]{x}\) and \(y=\frac{x}{4}\) that lies in the first quadrant about the y-axis.
Solution
Here are the graphs .The cross-sectional area is,
\(A= \pi [R(x)^2-r(x)^2] \)
or \(A= \pi [(4y)^2-(y^3)^2] \)
or \(A= \pi [16y^2-y^6] \)
Next, Working from the bottom of the solid to the top we can see that the first cross-section will occur at y=0 and the last cross-section will occur at y=2.
These will be the limits of integration.
Thus the volume of this solid is
\(V= \pi \int_a^b [R(x)^2-r(x)^2] dx \)
or \(V= \pi \int_a^b [16y^2-y^6] dx \)
or \(V=\pi \left ( \frac{16}{3} y^3 - \frac{1}{7} y^7 \right )|_0^2\)
or \(V=\frac{512 \pi}{21}\) - Determine the volume of the solid obtained by rotating the region bounded by y=x2-2x and y=x about the line y=4.
The radius of outer circle is
R(x)= \( 4-f(x)\)
or R(x)= \( 4-x^2+2x\)
The radius of inner circle is
r(x)= \( 4-g(x)\)
or R(x)= \( 4-x\)
Next, we need to determine the limits of integration.
The limits are x=0 and x=3.
Thus the volume of this solid is
\(V= \pi \int_a^b [R(x)^2-r(x)^2] dx \)
or \(V=\pi \int_{0}^3 \pi [(4-x^2+2x)^2-(4-x)^2] dx\)
or \(V=\frac{153 \pi}{5} \) - Find the volume of the object generated when the area between \(f(x)=x, g(x)=x^2-x\) rotated about the line \(y=3\).
- Find the volume of the object generated when the area between \(y=x^2,y=x\) rotated around the x-axis.
shell method
The shell method for finding volume of a solid uses integration along an axis perpendicular to the axis of revolution. Please note that disk and washer methods uses an axis parallel to the axis of revolution.
The shell method deals with an infinite number of “surface areas” of rectangles in the shapes of cylinders (shells) so that the volume of the cylindrical shell is
V=surface area of the cylinder x thickness of the cylindrical shell.
Let f(x) is non-negative and continuous on the interval [a,b], then the volume V formed by rotating the area under the curve of
f(x) about the y-axis is given as below.
For this suppose that,
location is x
height is h
thickness is dx
approximate radius is r
Tthen we cut the shell and laying it flat forms a rectangular solid with length 2πr, height h, and thickness dx
Then the volume of the cylindrical shell is
\(V≈(2πrh) dx\)
\(V=\displaystyle \sum_{i=1}^{n} (2 \pi r h) dx_i\)
\(V=\displaystyle \lim_{n \to \infty} \sum_{i=1}^{i=n} (2 \pi r h) dx_i\)
or \(V=\displaystyle \int_a^b (2 \pi r h) dx\)
Formuls for shell Method
- When the axis of rotation is the y -axis (i.e., x=0 then r=x )
- Volume= \(2 \pi \int_a^b x. f(x) dx \)
- When the axis of rotation is the x -axis (i.e., y=0 then r=y )
- Volume= \(2 \pi \int_a^b y. f( y) dy \)
- When the solid is bounded above by y=f(x) and below by y=g(x), then h=f(x)−g(x)
Solved Examples
- Determine the volume of the solid obtained by rotating the portion of the region bounded by \( y=\sqrt[3]{x}\) and \(y=\frac{x}{4}\) that lies in the first quadrant about the y-axis.
Solution
Here are the graphs . The volume of this solid is
\( V=2 \pi \int_0^8 x (x^{1/3}-0.25x) dx\)
Exercise
- Find the volume of the solid formed by rotating the region bounded by \(y=0 , y=\frac{1}{1+x^2},x=0 , x=1\) about the y -axis.
- Find the volume of the solid formed by rotating the triangular region determined by the points (0,1) , (1,1) and (1,3) about the line x=3
- Find the volume of the solid formed by rotating the triangular region determined by the points (0,1) , (1,1) and (1,3) about x-axis
- Find the volume of the solid formed by revolving the region bounded by y=sinx and the x-axis from x=0 to x=π about the y -axis.
- Find the volume of solid bounded by \(x=y^3,x=8\) and the x-axis and rotating the region about x-axis.
- Find the volume of solid bounded by \(f(x)=x+1,g(x)=(x-1)^2\) and the x-axis and rotating the region about y-axis.
- Find the volume of solid bounded by \(f(x\sqrt{x-1}+2\) rotating the region about y-axis between the lines x=1 and x=5.
- Suppose the area under \(y=-x^2+1\) is rotated around the x-axis. Find the volume of the solid of rotation.
- Suppose the area below the curve \(f(x)=x+1, x \in [0,3]\), is rotated about the y-axis. Find the volume
- Let R be the region of the xy-plane bounded below by the curve \(y=\sqrt{x}\) and, above by the line y=3. Find the volume of the solid obtained by rotating R around
- x-axis
- y-axis
- line y=3
- line x=9
- Let R be the region of the xy-plane bounded above by the curve \(x^3y=64\) and, below by the line y=1, on the left by the line x=2, and on the right by the line x=4. Find the volume of the solid obtained by rotating R around
- x-axis
- y-axis
- line y=1
- line x=2
- Use Shell Method to find the volumes of the solids generated by revolving the shaded region about the indicated axis.
- \(y=1+x^2\) about y-axis
- \(y=2-x^2\) about y-axis
- \(x=1+y^2\) about x-axis
- \(x=4-y^2\) about x-axis
- Use Shell Method to find the volumes of the solids generated by revolving the given bounded regions about the
y -axis.
- \(y=2x,y=-x,x=1\)
- \(y=x^2,y=x,y \ge 0\)
- \(y=x^2,y=x+2,x \ge 0\)
- \(y=1-x^2,y=x^2,x \ge 0\)
- \(y=\sqrt{x},y=2-x,x = 0\)
- \(y=\frac{1}{x},x=2,x=3,y=0\)
- Use Shell Method to find the volumes of the solids generated by revolving the given bounded regions about the
x -axis.
- \(x=2 \sqrt{y}, x=-y,y=4\)
- \(x=y^2, x=y, y \ge 0\)
- \(x=y-\frac{y^2}{4}, x=0\)
- \(x=y-\frac{y^2}{4}, x=\frac{y}{2}\)
- \(y=|x|,y=2\)
- \(y=x+1,y=2x,y=2\)
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