{"id":464,"date":"2025-02-13T19:45:13","date_gmt":"2025-02-13T19:45:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/physical-applications-learn-it-2\/"},"modified":"2025-02-13T19:45:13","modified_gmt":"2025-02-13T19:45:13","slug":"physical-applications-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/physical-applications-learn-it-2\/","title":{"raw":"Physical Applications: Learn It 2","rendered":"Physical Applications: Learn It 2"},"content":{"raw":"\n

Mass and Density Cont.<\/h2>\n

Mass\u2013Density Formula of a Two-Dimensional Disk<\/h3>\n

We now extend this concept to find the mass of a two-dimensional disk of radius [latex]r.[\/latex] As with the rod we looked at in the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as a two-dimensional object.<\/p>\n

We assume the density is given in terms of mass per unit area (called area density<\/em><\/span>), and further assume the density varies only along the disk\u2019s radius (called radial density<\/em><\/span>). We orient the disk in the [latex]xy[\/latex]-plane, with the center at the origin. Then, the density of the disk can be treated as a function of [latex]x,[\/latex] denoted [latex]\\rho (x).[\/latex] We assume [latex]\\rho (x)[\/latex] is integrable.<\/p>\n

Just as we did with the one-dimensional rod, we need to partition the interval. Partition the interval [latex][0,r][\/latex] into [latex]n[\/latex] segments, each of width [latex]\u0394x[\/latex].<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"613\"]\"This Figure 3. (a) A thin disk in the xy-plane. (b) A representative washer.[\/caption]\n\n

We now approximate the density and area of the washer to calculate an approximate mass, [latex]{m}_{i}.[\/latex] Note that the area of the washer is given by<\/p>\n

[latex]\\begin{array}{cc}\\hfill {A}_{i}& =\\pi {({x}_{i})}^{2}-\\pi {({x}_{i-1})}^{2}\\hfill \\\\ & =\\pi \\left[{x}_{i}^{2}-{x}_{i-1}^{2}\\right]\\hfill \\\\ & =\\pi ({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\\hfill \\\\ & =\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x.\\hfill \\end{array}[\/latex]<\/div>\n

You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use [latex]{x}_{i}^{*}\\approx ({x}_{i}+{x}_{i-1})\\text{\/}2[\/latex] to approximate the average radius of the washer. We obtain<\/p>\n

[latex]{A}_{i}=\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x\\approx 2\\pi {x}_{i}^{*}\\text{\u0394}x.[\/latex]<\/div>\n

Using [latex]\\rho ({x}_{i}^{*})[\/latex] to approximate the density of the washer, we approximate the mass of the washer by<\/p>\n

[latex]{m}_{i}\\approx 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n

Adding up the masses of the washers, we see the mass [latex]m[\/latex] of the entire disk is approximated by<\/p>\n

[latex]m=\\displaystyle\\sum_{i=1}^{n} {m}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n

We again recognize this as a Riemann sum, and take the limit as [latex]n\\to \\infty .[\/latex] This gives us<\/p>\n

[latex]m=\\underset{n\\to \\infty }{\\text{lim}} \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx.[\/latex]<\/div>\n

We summarize these findings in the following theorem.<\/p>\n

\n

mass\u2013density formula of a two-dimensional disk<\/h3>\n

Let [latex]\\rho (x)[\/latex] be an integrable function representing the radial density of a disk of radius [latex]r.[\/latex] Then the mass of the disk is given by<\/p>\n

[latex]m={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx[\/latex]<\/div>\n<\/section>\n
\n

Let [latex]\\rho (x)=\\sqrt{x}[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius [latex]4[\/latex].<\/p>\n\n[reveal-answer q=\"fs-id1167793477594\"]Show Solution[\/reveal-answer] [hidden-answer a=\"fs-id1167793477594\"]\n\n

Applying the formula, we find<\/p>\n

[latex]\\begin{array}{cc}\\hfill m& ={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{4}2\\pi x\\sqrt{x}dx=2\\pi {\\displaystyle\\int }_{0}^{4}{x}^{3\\text{\/}2}dx\\hfill \\\\ & =2\\pi {\\frac{2}{5}{x}^{5\\text{\/}2}|}_{0}^{4}=\\frac{4\\pi }{5}\\left[32\\right]=\\frac{128\\pi }{5}.\\hfill \\end{array}[\/latex]<\/div>\n
[\/hidden-answer]<\/div>\n<\/section>\n
\n
\n

[ohm_question hide_question_numbers=1]287495[\/ohm_question]<\/p>\n<\/section>\n<\/div>\n","rendered":"

Mass and Density Cont.<\/h2>\n

Mass\u2013Density Formula of a Two-Dimensional Disk<\/h3>\n

We now extend this concept to find the mass of a two-dimensional disk of radius [latex]r.[\/latex] As with the rod we looked at in the one-dimensional case, here we assume the disk is thin enough that, for mathematical purposes, we can treat it as a two-dimensional object.<\/p>\n

We assume the density is given in terms of mass per unit area (called area density<\/em><\/span>), and further assume the density varies only along the disk\u2019s radius (called radial density<\/em><\/span>). We orient the disk in the [latex]xy[\/latex]-plane, with the center at the origin. Then, the density of the disk can be treated as a function of [latex]x,[\/latex] denoted [latex]\\rho (x).[\/latex] We assume [latex]\\rho (x)[\/latex] is integrable.<\/p>\n

Just as we did with the one-dimensional rod, we need to partition the interval. Partition the interval [latex][0,r][\/latex] into [latex]n[\/latex] segments, each of width [latex]\u0394x[\/latex].<\/p>\n

\"This
Figure 3. (a) A thin disk in the xy-plane. (b) A representative washer.<\/figcaption><\/figure>\n

We now approximate the density and area of the washer to calculate an approximate mass, [latex]{m}_{i}.[\/latex] Note that the area of the washer is given by<\/p>\n

[latex]\\begin{array}{cc}\\hfill {A}_{i}& =\\pi {({x}_{i})}^{2}-\\pi {({x}_{i-1})}^{2}\\hfill \\\\ & =\\pi \\left[{x}_{i}^{2}-{x}_{i-1}^{2}\\right]\\hfill \\\\ & =\\pi ({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1})\\hfill \\\\ & =\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x.\\hfill \\end{array}[\/latex]<\/div>\n

You may recall that we had an expression similar to this when we were computing volumes by shells. As we did there, we use [latex]{x}_{i}^{*}\\approx ({x}_{i}+{x}_{i-1})\\text{\/}2[\/latex] to approximate the average radius of the washer. We obtain<\/p>\n

[latex]{A}_{i}=\\pi ({x}_{i}+{x}_{i-1})\\text{\u0394}x\\approx 2\\pi {x}_{i}^{*}\\text{\u0394}x.[\/latex]<\/div>\n

Using [latex]\\rho ({x}_{i}^{*})[\/latex] to approximate the density of the washer, we approximate the mass of the washer by<\/p>\n

[latex]{m}_{i}\\approx 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n

Adding up the masses of the washers, we see the mass [latex]m[\/latex] of the entire disk is approximated by<\/p>\n

[latex]m=\\displaystyle\\sum_{i=1}^{n} {m}_{i}\\approx \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x.[\/latex]<\/div>\n

We again recognize this as a Riemann sum, and take the limit as [latex]n\\to \\infty .[\/latex] This gives us<\/p>\n

[latex]m=\\underset{n\\to \\infty }{\\text{lim}} \\displaystyle\\sum_{i=1}^{n} 2\\pi {x}_{i}^{*}\\rho ({x}_{i}^{*})\\text{\u0394}x={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx.[\/latex]<\/div>\n

We summarize these findings in the following theorem.<\/p>\n

\n

mass\u2013density formula of a two-dimensional disk<\/h3>\n

Let [latex]\\rho (x)[\/latex] be an integrable function representing the radial density of a disk of radius [latex]r.[\/latex] Then the mass of the disk is given by<\/p>\n

[latex]m={\\displaystyle\\int }_{0}^{r}2\\pi x\\rho (x)dx[\/latex]<\/div>\n<\/section>\n
\n

Let [latex]\\rho (x)=\\sqrt{x}[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius [latex]4[\/latex].<\/p>\n