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200 g / 200 cm 3 = 1 g/cm 3; Optionally, change the unit. 1 g/cm 3 = 1 (1/1000 kg) / (1/1000000) m 3 = 1000 kg/m 3; Or you can use our density calculator to make it a breeze! The fastest way to find the density of an object is of course to use our density calculator. To make the calculation, you'll need to know a few other values to start with.
- I assume you mean 1.0 grams/milliliter. Pure water has a density of 1.0 g/ml.
- 1.1 Computing A measure of the amount of information on a storage medium (tape or disk). For magnetic tape it is the amount of information recorded per unit length of tape (bits per inch or millimeter); for a disk, a fixed number of bits per sector, sectors per track, and tracks per disk.
- Temperature o C: Density g/mL: Temperature o C: Density g/mL: 0: 0.99987: 15: 0.99913: 1: 0.99993: 20: 0.99823: 2: 0.99997: 25: 0.99707: 3: 0.99999: 30: 0.99562: 4: 1.
Mass of a Thin Rod
We can use integration for calculating mass based on a density function.
Consider a thin wire or rod that is located on an interval (left[ {a,b} right].)
Why Is 0 1
The density of the rod at any point (x) is defined by the density function (rho left( x right).) Assuming that (rho left( x right)) is an integrable function, the mass of the rod is given by the integral
[m = intlimits_a^b {rho left( x right)dx} .]
Mass of a Thin Disk
Suppose that (rho left( r right)) represents the radial density of a thin disk of radius (R.)
Then the mass of the disk is given by
[m = 2pi intlimits_0^R {rrho left( r right)dr} .]
Mass of a Region Bounded by Two Curves
Suppose a region is enclosed by two curves (y = fleft( x right),) (y = gleft( x right)) and by two vertical lines (x = a) and (x = b.)
If the density of the lamina which occupies the region only depends on the (x-)coordinate, the total mass of the lamina is given by the integral
[m = intlimits_a^b {rho left( x right)left[ {fleft( x right) – gleft( x right)} right]dx} ,]
where (fleft( x right) ge gleft( x right)) on the interval (left[ {a,b} right],) and ({rho left( x right)}) is the density of the material changing along the (x-)axis.
Mass of a Solid with One-Dimensional Density Function
Consider a solid (S) that extends in the (x-)direction from (x = a) to (x = b) with cross sectional area (Aleft( x right).)
Suppose that the density function (rho left( x right)) depends on (x) but is constant inside each cross section (Aleft( x right).)
The mass of the solid is
[m = intlimits_a^b {rho left( x right)Aleft( x right)dx} .]
Mass of a Solid of Revolution
Let (S) be a solid of revolution obtained by rotating the region under the curve (y = fleft( x right)) on the interval (left[ {a,b} right]) around the (x-)axis.
If (rho left( x right)) is the density of the solid material depending on the (x-)coordinate, then the mass of the solid can be calculated by the formula
[m = pi intlimits_a^b {rho left( x right){f^2}left( x right)dx} .]
Solved Problems
Click or tap a problem to see the solution.
Example 1
A rod with a linear density given by[rho left( x right) = {x^3} + x]lies on the (x-)axis between (x = 0) and (x = 2.) Find the mass of the rod.Example 2
Let a thin rod of length (L = 10,text{cm}) have its mass distributed according to the density function[rho left( x right) = 50{e^{ – frac{x}{{10}}}},]where (rho left( x right)) is measured in (large{frac{text{g}}{text{cm}}}normalsize,) (x) is measured in (text{cm}.) Calculate the total mass of the rod.Example 3
Suppose that the density of cars in traffic congestion on a highway changes linearly from 30 to 150 cars per km per lane on a (5,text{km}) long stretch. Estimate the total number of cars on the highway stretch if it has (4) lanes.Example 4
Determine the total amount of bacteria in a circular petri dish of radius (R) if the density at the center is ({rho_0}) and decreases linearly to zero at the edge of the dish.Example 5
Assuming that the stellar radial distribution within a galaxy obeys the exponential law[rho left( r right) = {rho _0}{e^{ – frac{r}{h}}},]estimate the mass of a galaxy with following parameters: ({rho _0} = {10^7}large{frac{{{M_{odot}}}}{{text{kpc}}}}normalsize,) (h = {10^4},text{kpc},) where ({M_{odot}}) denotes the solar mass and (text{kpc}) means a kiloparsec ((1,text{kpc} approx 3262) light-years).Example 6
A lamina occupies the region bounded by one arc of the sine curve and the (x-)axis. The density at any point of the lamina is proportional to the distance from the point to the (y-)axis. Find the mass of the lamina.Example 7
A lamina occupies the upper semicircle of radius (1) centered at the origin. Its density is given by the cubic function (rho left( y right) = {y^3}.) Find the mass of the lamina.Example 8
A right circular cone has base radius (R) and height (H.) What is the mass of the cone if its density varies along the vertical axis and is given by the function (rho left( y right) = k{y^2}?)Example 9
A right circular cone with base radius (R) and height (H) is formed by rotating about the (x-)axis. The density of the cone is given by the function (rho left( x right) = kx.) Find the mass of the cone assuming that the center of its base is placed in the origin.Example 10
The density of the Earth’s inner core is about (14200,large{frac{text{kg}}{text{m}^3}}normalsize,) and its average density near the surface is equal to (1160,large{frac{text{kg}}{text{m}^3}}normalsize.) Estimate the mass of the Earth if the density changes linearly and the Earth’s radius is (6370,text{km}.)Example 1.
A rod with a linear density given by[rho left( x right) = {x^3} + x]lies on the (x-)axis between (x = 0) and (x = 2.) Find the mass of the rod.Solution.
We need to integrate the following:
[{m = intlimits_a^b {rho left( x right)dx} }={ intlimits_0^2 {left( {{x^3} + x} right)dx} }={ left. {left( {frac{{{x^4}}}{4} + frac{{{x^2}}}{2}} right)} right|_0^2 }={ 6.}]
If (rho) is measured in kilograms per meter and (x) is measured in meters, then the mass is (m = 6,text{kg}.)
Example 2.
Let a thin rod of length (L = 10,text{cm}) have its mass distributed according to the density function[rho left( x right) = 50{e^{ – frac{x}{{10}}}},]where (rho left( x right)) is measured in (large{frac{text{g}}{text{cm}}}normalsize,) (x) is measured in (text{cm}.) Calculate the total mass of the rod.Solution.
To find the mass of the rod we integrate the density function:
[{m = intlimits_a^b {rho left( x right)dx} }={ intlimits_0^{10} {50{e^{ – frac{x}{{10}}}}dx} }={ – 500left. {{e^{ – frac{x}{{10}}}}} right|_0^{10} }={ 500left( {1 – frac{1}{e}} right) }={ frac{{500left( {e – 1} right)}}{e} }approx{ 316,text{g}.}]
Example 3.
Suppose that the density of cars in traffic congestion on a highway changes linearly from 30 to 150 cars per km per lane on a (5,text{km}) long stretch. Estimate the total number of cars on the highway stretch if it has (4) lanes.Solution.
First we derive the equation for the density function (rho left( x right).) Since the function is linear, it is defined by two points:
[{rho left( 0 right) = 30,;;}kern0pt{rho left( 5 right) = 150.}]
Using the two-point form of a straight line equation, we have
[{frac{{rho – 30}}{{150 – 30}} = frac{{x – 0}}{{5 – 0}},};; Rightarrow {frac{{rho – 30}}{{120}} = frac{x}{5},};; Rightarrow {rho – 30 = 24x,};; Rightarrow {rho left( x right) = 24x + 30.}]
Now, to estimate the amount of cars on the highway stretch, we integrate the density function and multiply the result by (4:)
[{N = 4intlimits_a^b {rho left( x right)dx} }={ 4intlimits_0^5 {left( {24x + 30} right)dx} }={ left. {4left( {12{x^2} + 30x} right)} right|_0^5 }={4left({ 300 + 150 }right) }={ 1800,text{cars}}.]
Example 4.
Determine the total amount of bacteria in a circular petri dish of radius (R) if the density at the center is ({rho_0}) and decreases linearly to zero at the edge of the dish.Solution.
The density of bacteria varies according to the law
[rho left( r right) = {rho _0}left( {1 – frac{r}{R}} right),]
where (0 le r le R.)
To find the total number of bacteria in the dish, we use the formula
[N = 2pi intlimits_0^R {rrho left( r right)dr} .]
This yields
[{N = 2pi {rho _0}intlimits_0^R {rleft( {1 – frac{r}{R}} right)dr} }={ 2pi {rho _0}intlimits_0^R {left( {r – frac{{{r^2}}}{R}} right)dr} }={ 2pi {rho _0}left. {left( {frac{{{r^2}}}{2} – frac{{{r^3}}}{{3R}}} right)} right|_0^R }={ 2pi {rho _0}left( {frac{{{R^2}}}{2} – frac{{{R^2}}}{3}} right) }={ frac{{2pi {rho _0}{R^2}}}{6} }={ frac{{pi {rho _0}{R^2}}}{3}.}]
Example 5.
Assuming that the stellar radial distribution within a galaxy obeys the exponential law[rho left( r right) = {rho _0}{e^{ – frac{r}{h}}},]estimate the mass of a galaxy with following parameters: ({rho _0} = {10^3}large{frac{{{M_{odot}}}}{{{text{kpc}^2}}}}normalsize,) (h = {10^4},text{kpc},) where ({M_{odot}}) denotes the solar mass and (text{kpc}) means a kiloparsec ((1,text{kpc} approx 3262) light-years).Solution.
We will assume that the galaxy has the form of a thin disc and therefore it is possible to apply the formula
[m = 2pi intlimits_0^R r rho left( r right)dr.]
Since the exact value of the radius (R) of the galaxy is unknown, we will set it equal to infinity. Calculate the improper integral:
[{m = 2pi intlimits_0^infty {rrho left( r right)dr} }={ 2pi {rho _0}intlimits_0^infty {r{e^{ – frac{r}{h}}}dr} }={ 2pi {rho _0}lim limits_{b to infty } intlimits_0^b {r{e^{ – frac{r}{h}}}dr}.}]
Integrating by parts, we obtain:
[{int {underbrace r_uunderbrace {{e^{ – frac{r}{h}}}dr}_{dv}} }={ left[ {begin{array}{*{20}{l}}
{u = r}
{dv = {e^{ – frac{r}{h}}}dr}
{du = dr}
{v = – h{e^{ – frac{r}{h}}}}
end{array}} right] }={ – hr{e^{ – frac{r}{h}}} – int {left( { – h{e^{ – frac{r}{h}}}} right)dr} }={ – hr{e^{ – frac{r}{h}}} + hint {{e^{ – frac{r}{h}}}dr} }={ – hr{e^{ – frac{r}{h}}} – {h^2}{e^{ – frac{r}{h}}} }={ – hleft( {r + h} right){e^{ – frac{r}{h}}}.}]
{u = r}
{dv = {e^{ – frac{r}{h}}}dr}
{du = dr}
{v = – h{e^{ – frac{r}{h}}}}
end{array}} right] }={ – hr{e^{ – frac{r}{h}}} – int {left( { – h{e^{ – frac{r}{h}}}} right)dr} }={ – hr{e^{ – frac{r}{h}}} + hint {{e^{ – frac{r}{h}}}dr} }={ – hr{e^{ – frac{r}{h}}} – {h^2}{e^{ – frac{r}{h}}} }={ – hleft( {r + h} right){e^{ – frac{r}{h}}}.}]
Taking limits then yields:
[{m = 2pi {rho _0}limlimits_{b to infty } intlimits_0^b {r{e^{ – frac{r}{h}}}dr} }={ 2pi {rho _0}hlimlimits_{b to infty } left[ {left. {left( { – left( {r + h} right){e^{ – frac{r}{h}}}} right)} right|_0^b} right] }={ 2pi {rho _0}hlimlimits_{b to infty } left[ {h – frac{{b + h}}{{{e^{frac{b}{h}}}}}} right].}]
1 0 Math
By L’Hopital’s rule we have that Graphicriver brochure download free.
[{limlimits_{b to infty } left[ {h – frac{{b + h}}{{{e^{frac{b}{h}}}}}} right] }={ h – limlimits_{b to infty } frac{{left( {b + h} right)^prime}}{{left( {{e^{frac{b}{h}}}} right)^prime}} }={ h – limlimits_{b to infty } frac{0}{{frac{1}{h}{e^{frac{b}{h}}}}} }={ h.}]
Hence, the mass of the galaxy is given by the equation
[m = 2pi {rho _0}{h^2}.]
Substitute the given values:
[{m = 2pi times {10^3} times {left( {{{10}^4}} right)^2} }={ 2pi times {10^{11}} }approx{ 6.28 times {10^{11}},{M_odot}},]
that is about (2) times less than the mass of the Milky Way.
Example 6.
A lamina occupies the region bounded by one arc of the sine curve and the (x-)axis. The density at any point of the lamina is proportional to the distance from the point to the (y-)axis. Find the mass of the lamina.Solution.
The general formula for the mass of a region between two curves is
[m = intlimits_a^b {rho left( x right)left[ {fleft( x right) – gleft( x right)} right]dx} .]
Substituting the known functions and limits, we get:
[m = intlimits_0^pi {xsin xdx} .]
To evaluate the integral we use integration by parts:
[{m = intlimits_0^pi {underbrace x_uunderbrace {sin xdx}_{dv}} }={ left[ {begin{array}{*{20}{l}}{u = x}{dv = sin xdx}{du = dx}{v = – cos x}end{array}} right] }={ left. {left( { – xcos x} right)} right|_0^pi – intlimits_0^pi {left( { – cos x} right)dx} }={ left. {left( { – xcos x} right)} right|_0^pi + intlimits_0^pi {cos xdx} }={ left. {left( {sin x – xcos x} right)} right|_0^pi }={ pi .}]
If the density ({rho left( x right)}) is measured in (large{frac{text{kg}}{text{m}}}normalsize) and (x) is measured in (text{m},) then the mass of the lamina is (m = pi,text{kg}.)
Example 7.
A lamina occupies the upper semicircle of radius (1) centered at the origin. Its density is given by the cubic function (rho left( y right) = {y^3}.) Find the mass of the lamina.Solution.
Here the density function varies along the (y-)axis. Therefore we use the following formula to calculate the mass of the lamina:
[m = intlimits_c^d {rho left( y right)left[ {fleft( y right) – gleft( y right)} right]dy} .]
Due to symmetry about the (y-)axis, we can integrate from (0) to (1) and then multiply the answer by (2.)
The circle in the first quadrant is given by the equation (x = fleft( y right) = sqrt {1 – {y^2}} .) Hence, the mass of the lamina is expressed by the integral
[m = 2intlimits_0^1 {{y^3}sqrt {1 – {y^2}} dy} .]
To evaluate the integral, we change the variable:
[{1 – {y^2} = {z^2},};; Rightarrow {{y^2} = 1 – {z^2},;;}kern0pt{ydy = – zdz,;;}kern0pt{{y^3}dy = left( {1 – {z^2}} right)left( { – zdz} right) }={ left( {{z^3} – z} right)dz.}]
When (y = 0,) (z = 1,) and when (y = 1,) (z = 0.) So, the integral in terms of (z) is written as
[{m = 2intlimits_1^0 {left( {{z^3} – z} right)zdz} }={ 2intlimits_1^0 {left( {{z^4} – {z^2}} right)dz} }={ 2left. {left[ {frac{{{z^5}}}{5} – frac{{{z^3}}}{3}} right]} right|_1^0 }={ 2left[ {0 – left( {frac{1}{5} – frac{1}{3}} right)} right] }={ 2left( {frac{1}{3} – frac{1}{5}} right) }={ frac{4}{{15}}.}]
Example 8.
A right circular cone has base radius (R) and height (H.) What is the mass of the cone if its density varies along the vertical axis and is given by the function (rho left( y right) = k{y^2}?)Solution.
Consider a thin slice at a height (y) parallel to the base. The radius (r) of the cross section can be determined from the proportion for similar triangles:
[{frac{r}{R} = frac{{H – y}}{H},};; Rightarrow {r = frac{R}{H}left( {H – y} right) .}]
The mass of the slice of thickness (dy) is given by
[{dm = rho left( y right)dV = pi {r^2}rho left( y right)dy }={ frac{{pi {R^2}rho left( y right){{left( {H – y} right)}^2}}}{{{H^2}}}dy }={ frac{{pi k{R^2}{y^2}{{left( {H – y} right)}^2}}}{{{H^2}}}dy.}]
To compute the total mass of the cone, we integrate from (y = 0) to (y = H:)
[{m = frac{{pi k{R^2}}}{{{H^2}}}intlimits_0^H {{y^2}{{left( {H – y} right)}^2}dy} }={ frac{{pi k{R^2}}}{{{H^2}}}intlimits_0^H {{y^2}left( {{H^2} – 2Hy + {y^2}} right)dy} }={ frac{{pi k{R^2}}}{{{H^2}}}intlimits_0^H {left( {{H^2}{y^2} – 2H{y^3} + {y^4}} right)dy} }={ frac{{pi k{R^2}}}{{{H^2}}}left. {left( {frac{{{H^2}{y^3}}}{3} – frac{{H{y^4}}}{2} + frac{{{y^5}}}{5}} right)} right|_0^H }={ pi k{R^2}{H^3}left( {frac{1}{3} – frac{1}{2} + frac{1}{5}} right) }={ frac{{pi k{R^2}{H^3}}}{{30}}.}]
Let’s check the answer using dimensional analysis. The density is expressed by the function (rho left( y right) = k{y^2}.) So, if the variable (y) is measured in metres and the mass is measured in kilograms, then the coefficient (k) is measured in (large{frac{text{kg}}{text{m}^5}}normalsize.) Hence,
[require{cancel}{{m} = {frac{{pi k{R^2}{H^3}}}{{30}}}}={ left[ {frac{{kg}}{{{m^5}}}} right]left[ {{m^2}} right]left[ {{m^3}} right] }={ frac{{left[ {kg} right]cancel{left[ {{m^5}} right]}}}{{cancel{left[ {{m^5}} right]}}} }={ left[ {kg} right].}]
Example 9.
A right circular cone with base radius (R) and height (H) is formed by rotating about the (x-)axis. The density of the cone is given by the function (rho left( x right) = kx.) Find the mass of the cone assuming that the center of its base is placed in the origin.Solution.
We calculate the mass of the cone by the formula
[m = piintlimits_a^b {rho left( x right){f^2}left( x right)dx} .]
The equation of the straight line (y = fleft( x right)) is expressed as follows:
[{y = fleft( x right) }={ R – frac{R}{H}x }={ frac{R}{H}left( {H – x} right).}]
Substituting the density function (rho left( x right) = kx) and integrating from (x = 0) to (x = H), we have
[{m = pi intlimits_0^H {kxfrac{{{R^2}}}{{{H^2}}}{{left( {H – x} right)}^2}dx} }={ frac{{pi k{R^2}}}{{{H^2}}}intlimits_0^H {x{{left( {H – x} right)}^2}dx} }={ frac{{pi k{R^2}}}{{{H^2}}}intlimits_0^H {xleft( {{H^2} – 2Hx + {x^2}} right)dx} }={ frac{{pi k{R^2}}}{{{H^2}}}intlimits_0^H {left( {{H^2}x – 2H{x^2} + {x^3}} right)dx} }={ frac{{pi k{R^2}}}{{{H^2}}}left. {left( {frac{{{H^2}{x^2}}}{2} – frac{{2H{x^3}}}{3} + frac{{{x^4}}}{4}} right)} right|_0^H }={ pi k{R^2}{H^2}left( {frac{1}{2} – frac{2}{3} + frac{1}{4}} right) }={ frac{{5pi k{R^2}{H^2}}}{{12}}.}]
Note that if the cone’s radius and height are measured in metres and the mass in kilograms, then the coefficient (k) is measured in ({large{frac{{text{kg}}}{{{text{m}^4}}}}normalsize}.)
Proof That 1 Equals 0
Example 10.
The density of the Earth’s inner core is about (14200,large{frac{text{kg}}{text{m}^3}}normalsize,) and its average density near the surface is equal to (1160,large{frac{text{kg}}{text{m}^3}}normalsize.) Estimate the mass of the Earth if the density changes linearly and the Earth’s radius is (6370,text{km}.)Solution.
First, let’s derive the equation for calculating the mass of a ball with a linear density distribution.
If we take an arbitrary thin layer of thickness (dr) at a distance (r) from the center, its volume is given by
[dV = 4pi {r^2}dr.]
The mass of the layer is
[{dm = rho left( r right)dV }={ 4pi rho left( r right){r^2}dr.}]
Hence, the total mass of the ball is given by the integral
[m = 4pi intlimits_0^R {rho left( r right){r^2}dr} .]
Assuming that the density decreases linearly, we write it in the form (rho left( r right) = a – br,) where (a) and (b) are positive coefficients that can be found from the boundary conditions. Then the mass of the Earth is expressed as follows:
[{m = 4pi intlimits_0^R {left( {a – br} right){r^2}dr} }={ 4pi intlimits_0^R {left( {a{r^2} – b{r^3}} right)dr} }={ 4pi left. {left( {frac{{a{r^3}}}{3} – frac{{b{r^4}}}{4}} right)} right|_0^R }={ 4pi left( {frac{{a{R^3}}}{3} – frac{{b{R^4}}}{4}} right).}]
Determine the coefficients (a) and (b.) Given that
[{rho left( 0 right) = 14200frac{{text{kg}}}{{{text{m}^3}}},;;}kern0pt{rho left( R right) = 1160frac{{text{kg}}}{{{text{m}^3}}},;;}kern0pt{R = 6370,text{km} = 6.37times 10^6,text{m,}}]
Temp monitor 1 0 8 download free. and using the two-point form of a straight line equation, we get
[{frac{{rho left( r right) – rho left( 0 right)}}{{rho left( R right) – rho left( 0 right)}} = frac{{r – 0}}{{R – 0}},};; Rightarrow {rho left( r right) = rho left( 0 right) + frac{{rho left( R right) – rho left( 0 right)}}{R}r }={ 14200 + frac{{1160 – 14200}}{{6.37 times {{10}^6}}}r }={ 14200 – 2.047 times {10^{ – 3}}r.}]
So, (a = 14200) and (b = 2.047times{10^{-3}}.)
Now we can compute the total mass of the Earth:
[{m = 4pi left( {frac{{a{R^3}}}{3} – frac{{b{R^4}}}{4}} right) }={ 4pi left[ {frac{{1.42 times {{10}^4} times {{left( {6.37 times {{10}^6}} right)}^3}}}{3} }right.}-{left.{ frac{{2.047 times {{10}^{ – 3}} times {{left( {6.37 times {{10}^6}} right)}^4}}}{4}} right] }={ 4pi left[ {122.34 times {{10}^{22}} – 842.59 times {{10}^{21}}} right] }approx{ 4.79 times {10^{24}},text{kg}.}]
This result is about (20%) less than the actual Earth’s mass, which is equal to (5.97 times {10^{24}},text{kg}.) This means that the inner layers are actually more dense than the linear approximation suggests.